General Physics General Physics Help Forum 
Jun 17th 2014, 07:27 PM

#11  Junior Member
Join Date: May 2014
Posts: 20
 Gravity Generator
So if you’re game, here’s where it gets complicated….So let’s say you have 1338 cubic foot pressure tank and you have added 3315 cubic feet of air into it. The tank is now at 2.47 atmospheres or right around 36 psi. So now we want to add 255 more cubic feet of air into the tank. But instead of using a motorized compressor, we want to use a gravity compressor of sorts.
Let’s say you have a mass that weighs 15,329 lbs. or 6,953.11 Kg and you have ten feet of fall to play with. Between the air intake of the tank and the weight, you have either one very large piston, or a series of smaller pitons. The chamber(s) of the one piston or multiple pistons shall equal 255 cubic feet. You will be using the weight to drive the piston and force the air into the pressure tank (which is already exerting 36 psi against you). You can use the 10 foot fall any way you wish. You can have the weight drop in ten 1 foot increments or 2 five foot drops, or one long 10 foot drop whichever is optimal. You could have the weight free fall for 5 or 9 feet before hitting the piston adding additional force.
I would like to know how much of the 255 cubic feet of air could you force into that pressure tank given these parameters to work with. Also I would like to know if you changed the size of the tank, say you made it half the size: 669 cubic feet and you add 1657.5 cubic feet to get your 2.47 atmospheres and 36 PSI. Would the amount of air you could force into the vessel increase, decrease or remain the same as in the previous example?

 
Jun 17th 2014, 08:38 PM

#12  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,320

Given 255 ft^3 of air and a piston 10ft long, the area lf the piston would be 25.5 ft^2. If the piston has a weight of 15,339 pounds driving it, that's 15329/25.5 = 601 psi of pressure it exerts.
** EDIT  This is incorrect  it needs to be converted from pounds/sq ft to pounds/sq inch. See post #14 below. **
It would have no problem driving against the 36 PSI resistance of the air already in the tank. Consequently, it can force the full 255 ft^3 of air into the tank, raising its pressure another 0.19 atmospheres.
If the tank is half the size but still starting at 36 PSI, the piston would again have no problem driving an additional 255 ft^3 of air into it, although the pressure increase for each shot of air would be 2x0.19 = 0.38 atmospheres.
Last edited by ChipB; Jun 18th 2014 at 05:34 AM.

 
Jun 17th 2014, 11:44 PM

#13  Junior Member
Join Date: May 2014
Posts: 20

That was way better then I hoped. I'd like to modify the question. Let's say I doubled the amount I wanted to add to the tank by adding 510 cubic feet. I would like to know:
A) What would be the least amount of weight necessary to push that
(2x) piston, and
B) Is there a formula to predict the rate of descent e.g If the weight were
10,000 pounds, at what rate would the pistons fall?

 
Jun 18th 2014, 05:33 AM

#14  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,320

I realize I made a big mistake in my last post. Apologies, but I should have converted ft^2 to in^2 to get the correct answer. The 15329 weight can exert a pressure of:
15329/25.5ft^2 x 1 ft^2/144in^2 = 4.18 PSI.
So unfortunately the weight can not compress any more air into the container if it's already at 36 PSI. You're going to have to be satisfied with a smaller area of the piston, which means either (a) less than 255 ft^3 of air being compressed, or (b) a longer stroke of the piston. To get the pressure the piston creates up to 36 PSI while limited to a tenfoot stroke its area must be (4.18/36) x 25.5 = 2.9 ft^2, and hence the volume of air is 2.9 ft^2 x 10ft = 29 ft^3.
Last edited by ChipB; Jun 18th 2014 at 09:46 AM.

 
Jun 19th 2014, 10:42 PM

#15  Junior Member
Join Date: May 2014
Posts: 20
 Yea, that first answer made things way too easy. It took me a few days, but I believe I have discovered a work around which leads me to another question. I did the best I could to use your formulas, but if you could confirm my calculations that would be great.Let’s say you have a 4500 gallon pressure tank or 602 cu/ft. There is 825 cu/ft of air in the tank which puts the pressure at 20.13 PSI. I want to add 300 cu/ft of air to the tank which will bring the pressure up to 27.48 PSI.Now let’s say that I have a piston which weighs 15329 LBS and is 26 inches in diameter. This means the piston should produce 28.87 PSI when used as a “gravity compressor”. Presuming these numbers are correct, the question is this: You have the pressure tank at 20.13 PSI. Coming out of the pressure tank you have a cylinder 26 inches in diameter that stands 100 feet high. The volume of air in the cylinder should be right at 307 cu/ft. At the top of the cylinder is your 15329 LBS piston. If you were to drop the piston, how long would it take to fall the 100 feet and compress all of the 307 cu/ft of air into the tank?Would the fall of the piston have to be controlled? In other words would there be so much momentum from the weight that it would hit and damage the tank or would the pressure building up in the cylinder be able to control the descent of the piston. Is there a formula to calculate this which I can use to play around with different weights and heights?Cheers,Kenji
Last edited by tonk; Jun 19th 2014 at 10:44 PM.

 
Jun 20th 2014, 10:53 AM

#16  Senior Member
Join Date: Apr 2008 Location: Bedford, England
Posts: 668

The Combined Gas Law looks like a good starting point: http://en.wikipedia.org/wiki/Combined_gas_law
Note the Temperature term, this means the gas will get hot as it is compressed.
This means you will have to thermally insulate the cylinder to maintain efficiency.

 
Jun 21st 2014, 09:36 AM

#17  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,320

I am traveling and unable to provide a complete answer to your questions about time to fall 100 ft and the speed of impact. But the general approach is this:
1. Recognizing MBW's point about the air heating under compression, this will serve to raise the pressure above what the simple volume change calculation implies. However the math becomes very difficult, so instead let's assume that the walls of the cylinder are not well insulated, and that heat is allowed to escape so that the temp of the air stays constant. This will give a first order approximation, which would tend to underestimate the time and overestimate the final velocity.
2. The equation of motion to use is
Integral adx = integral vdv
Where a is acceleration, and equals sum of forces/m. The forces involved are gravity and the opposing pressure. Assuming a linear build up of pressure as the weight descends, you have P(x)= P_1 + Delta p (x/L), where P_1 is the initial pressure and Delta P is the change in pressure as x goes from 0 to 100 ft. So the equation becomes:
(Mgx P_1Ax Delta P Ax^2/2)/m = v^2/2
This yields an equation for v(x), and from that the final velocity is v(L) and the time to reach the bottom is found from integral of integral of adt from t= 0 to t= T is equal to v(L), so you can solve for T.
I suggest you try to solve this yourself. Otherwise give me a day or two to get back on this.

 
Jan 2nd 2015, 11:33 PM

#18  Junior Member
Join Date: May 2014
Posts: 20

Greetings, It's been awhile. I have a question on water compressibility. Say you had an unpressurized 1000 gallon water tank completely full of water and you wanted to force 2 more gallon of water into the tank. How much pressure would be needed? What if it were 10000 gallons and you want to force in 5 gallons. I don't need the specific answers, I'm looking for a formula in which I can plug in different values. Thank you.

 
Jan 3rd 2015, 07:10 AM

#19  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,320

The general formula for compressing water (or anything else) into a given volume is:
s=Ee
where s (sigma) is the stress needed to be applied in terms of pressure (in units such as psi or Pa), E is the bulk modulus of the material, and e (epsilon) is the strain you want to achieve, in units of change of length per unit length. For water E = 2.2 x 10^9 Gpa. The strain you are wanting to achieve is equal to the added volume divided by the original volume, so for your first example it's 2/1000. Hence the pressure you need to apply is:
E= 2.2 x 10^9 Pa x (2/1000) = 4.4x10^6 Pa, or 4.4x10^6 N/m^2, which is equivalent to about 640 psi.

 
Jan 3rd 2015, 10:35 AM

#20  Junior Member
Join Date: May 2014
Posts: 20

Thank you. I am assuming gases are much easier to compress. Could you tell me which gases compress the most easily. 640 psi is to high for what I am trying to do and I would like to mitigate that value. Using the same example of the 2 gallons into a 1000 gallon tank, what if inside that water tank there was a pliable sphere filled with air (or the best compressible gas) with a radius of 6 inches. If my math is correct, the sphere would have a volume of .523599 cu/ft or just under 4 gallons. Now how much pressure would you need to force those 2 gallons into the tank? How would the size of the sphere affect the pressure needed to insert the 2 gallons into the tank, in other words would larger spheres make it easier? I would also need the formula. I was also wondering, are you allowed to or available to consult on projects? Thanks Kenji

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