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Offline spark727

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Need some help - seeking equation
« on: 12/05/2011 02:21:30 »
Hi, I want to lift up a person 10m of the ground (solid, not water) in 10 seconds, the person has a weight including equipment of 120Kg, the lifting force will be just regular air (mainstreem air), produced by an airpump. - now the question is: what will be the PRESSURE (bar), and how much Airflow (m3) from this pump. (please give me also an equation )
Additional Details
First, it would be a cylinder of compressed air that expands to raise the person upwards, and this cylinder is tied up to the persons back.
Secondly the piston (or bottle output) diameter would be 200mm.
the air is just regular air (mainstreem air), NOT a heated yet air. :-\



 

Offline yor_on

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Need some help - seeking equation
« Reply #1 on: 12/05/2011 17:20:14 »
I think it is this you're looking for?

    PV = nRT


Where P is pressure. V is volume. T is temperature (Kelvin). R is the universal gas constant (8.3145 J/mol K) and n is the number of moles of gas molecules (mole volume).

To calculate the pressure, calculate the number of moles n in your gas.
Multiply that by R, multiply by temperature, divide by volume.

You don't need the surface area, but you need to know the volume.

Knowing the mass of the gas you can divide it by the molar mass.
Knowing the number of molecules you can divide by Avogadro’s number (number of atoms in 1 mole). And air seems to fit this rather well, as I understands it, even though being a mix.

Take a look here Ideal gas.
Is that a home work question? Or is it a serious idea, kind of unusual if so :)


==

And a "mole is defined as the amount of substance that contains as many elementary entities (e.g., atoms, molecules, ions, electrons) as there are atoms in 12 g of the isotope carbon-12. Thus, by definition, one mole of pure 12C has a mass of exactly 12 g.

The experimentally determined value of a mole as adopted by CODATA in 2006 is 6.02214179(30)×1023

By this definition, a mole of any pure substance has a mass in grams exactly equal to that substance's molecular or atomic mass; e.g., 1mol of calcium-40 has an approximative mass of 40g, because the Ca-40 isotope has a mass of 39.9625906 amu on the C-12 scale. In other words, the numerical value of a substance's molecular or atomic mass in atomic mass units is the same as that of its molar mass—the mass of one mole of that substance—in grams.

The most common method of determining the amount, expressed in moles, of pure substance the value of whose molar mass is known, is to measure its mass in grams and then to divide by its molar mass (expressed in g/mol).[8] Molar masses may be easily calculated from tabulated values of atomic weights and the molar mass constant (which has a convenient defined value of 1 g/mol). Other methods include the use of the molar volume or the measurement of electric charge."
« Last Edit: 12/05/2011 17:24:17 by yor_on »
 

Offline Soul Surfer

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Need some help - seeking equation
« Reply #2 on: 12/05/2011 21:42:31 »
The answer to this question is not single valued.  It depends on the velocity with which the air is expelled it is essentially a rocket equation.
 

Offline spark727

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« Reply #3 on: 12/05/2011 22:18:43 »
no, this is not a homework, this is the real deal.
 

Offline rosy

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« Reply #4 on: 12/05/2011 22:45:13 »
Quote
no, this is not a homework, this is the real deal.

Hang on... You're actually planning to fire someone off the ground up to (more-or-less) the roof level of a three storey building, and you're asking people you've never met and know nothing about to provide you with "an equation" to calculate how you would do this? Even if I had a clear picture of what sort of mechanism you're planning to use (which from your original question I don't), I don't think it would be responsible to provide you with the information you're asking for because I think it would be encouraging you in what looks to me to be a dangerous and ill-thought-out enterprise.

Or have I misunderstood, are you trying to construct some form of elevator with the pump forcing air into a tube and the rider being raised up the tube on a platform?

If you want an answer, in any case, you'll have to phrase your question rather more clearly.
 

Offline CliffordK

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Need some help - seeking equation
« Reply #5 on: 12/05/2011 22:46:41 »
The answer to this question is not single valued.  It depends on the velocity with which the air is expelled it is essentially a rocket equation.
Is it rockets or hydraulics?

I.E.
Are you pushing a piston in a closed system, or a jet/rocket engine in an open system.  Both giving very different results.

In a closed (piston) system you can use a compressible gas such as air.  It has the advantage that you can fill a storage tank (compressor tank) and quickly transfer it to your cylinder.  

However, there are also advantages of using oils as in hydraulic cylinders.

Elevators or forklifts might also use chain drives or cable drives.

Some kind of a strap-on rocket pack is great for movies, but very dangerous in real life.
 

Offline spark727

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Need some help - seeking equation
« Reply #6 on: 12/05/2011 22:48:31 »
Let me ask you guys again.

1.) I want to lift up a person including equipment 10m of the soild ground in 10 seconds.
2.) The persons weight (including equipment) is 120Kg
3.) The propoulsion will be press'd air (NOT hot turbine air), just by a conventional air pump.
4.) The presure output, will be 2 nossles, wit a diameter of 100mm each.

Now the question is:
A.) How many Bar air pressue of the air pump is required?
B.) How many cubic meter of air will be the output of the air pump? ???
                      
 

Offline spark727

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Need some help - seeking equation
« Reply #7 on: 12/05/2011 22:54:08 »
NO guys, this is a serious question, and qhat is so wrong with it??

If i would a asked you give me a rocket eqation to propell this weight up in air, i would a got it with the first answer (i believe).

Now what i am asking, is NOT famable, is just regular narmal (mainstreem) air, and we getting questions??

WHY???

because there is no such technology availabe??? :-\
 

Offline spark727

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Need some help - seeking equation
« Reply #8 on: 12/05/2011 22:56:55 »
no CliffodK,
this is just a regular air pump - like you pond pump you know, just with air, and not with water.
 

Offline spark727

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Need some help - seeking equation
« Reply #9 on: 12/05/2011 23:08:25 »
Rosy
"Or have I misunderstood, are you trying to construct some form of elevator with the pump forcing air into a tube and the rider being raised up the tube on a platform?"

Yes Rosy, you can say so. 
 

Offline CliffordK

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« Reply #10 on: 13/05/2011 01:01:55 »
Ok,
So this is essentially rockets and thrust (but without combustion).  Sorry I don't have the equations on hand, but it would be the same as any thrust equation.

Your volume of air required to provide lift will be huge.  Think about moving about 300 lbs worth of air 9.8 meters every second.  Obviously faster moving air means less volume.  But, if you are just doing air & fans, think of using some big fans.  Perhaps 2 or 3 fans a meter in diameter.




If you are only wanting a 10-second flight. 

Consider taking an aluminum or composite compressed gas cylinder (air cylinder) capable of holding a few thousand PSI, and releasing the gas very quickly.

One is supposed to get an ordinary welding cylinder to fly quite nicely by knocking the valve off.
 

Offline JP

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Need some help - seeking equation
« Reply #11 on: 13/05/2011 04:32:08 »
This should be basically what you'rel looking for:

http://en.wikipedia.org/wiki/Impulse_(physics)

You can calculate the force required to lift the person, and then you know the change in momentum required.  The gas discharged carries momentum equal to its mass times its velocity.  By conservation of momentum that is imparted to the person being lifted.  The rate of gas discharge has to give a dp/dt to give you the proper force of lift: you can achieve this by adjusting velocity or mass of the gas discharged per second. 

Of course the real world situation won't be perfectly efficient, so you'll need more gas or higher velocity in practice.
 

Offline yor_on

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Need some help - seeking equation
« Reply #12 on: 14/05/2011 04:04:41 »
Ah. you wouldn't happen to be one of those daredevil Australians by any chance :)
I gotta admit that it's a lovely way to go :) All for science.

But sweaty, quite sweaty, and your insurance company won't like it. On that I'm reasonably sure :)
==

By the way, I read it as you wanting to use some sort of gas filled contraption, not reaction driven and self propelled.
« Last Edit: 14/05/2011 04:07:42 by yor_on »
 

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Need some help - seeking equation
« Reply #12 on: 14/05/2011 04:04:41 »

 

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