# Lookin for a formula

1. Aug 9, 2010

### VA80232

This doesn't have to be exact. so if a some what simple formula might do the job, great.

I am requesting a formula for:

If I had a 100 ft tall cylinder. 2ft wide and 2ft long at the top and bottom equally. A square cylinder. I dropped a 25 pound weight from the top and it free fell loosing no air, at sea level.

How far would it drop towards the bottom before it stopped and held its own weight and how many pounds pressure would the compressed air be left in the cylinder?

This is not homework I am a vet senior citizen with a curiosity. Thank you for your help ahead of time.

2. Aug 9, 2010

### Mech_Engineer

I'm assuming you're talking about a float in water...

It all depends on how much the "cylinder" weighs. Basically all you need to do is calculate the volume of water being displaced an multiply by the density of water. His will give you the buoyancy of the float; subtract the weight of the system and you'll either have a net buoyancy (in which case it floats) or net weight (sinks).

3. Aug 10, 2010

### litup

I think we need some more information on your problem. For one thing, you didn't say if the top has a lid or the bottom has a lid or both or neither. The other guy answering seems to think you are IN the sea when you specified only the cylinder was at sea LEVEL which I assume you mean the cylinder is on land not water so will not be sinking. So does the cylinder have a lid or two or none? If no lid then you would loose air over the top, some air, depending on the size and shape of the 25 pound weight. Also, what is the shape of the weight? Is it like a washer, where the edge of the weight almost touches the inner walls? Or is it a sphere, like a big ball bearing? All those things make a difference when you construct problems like this. You need to specify more detail. So reconstruct your problem based on this advice, ok?

For instance, the density of the weight also matters, is it lead? Is it styrofoam? Do you see what difference that kind of thing can make? 25 pounds of lead, if it were in a shape of a ball, would be only a few inches across and wouldn't effect the air flow as much as if it were a plug of wood just under 2 feet square and a few inches thick. There, the air would slow down the plug because it would try to compress the air underneath and whatever air slips by the gaps at the edge would tell how much time goes by before it hits the bottom. If it were totally sealed but free to fall for instance, it would be basically a weight adding 25 pounds to the air column, compressing the air by some amount and altering the pressure inside (also the temperature a little bit), kind of like a pneumatic shock absorber, that's what happens inside it, it just compresses air which acts as a damper.

So what is it here?

Last edited: Aug 10, 2010
4. Aug 10, 2010

### minger

OK, let me take a stab at what I 'think' you're talking about. Let's assume that you have a hollow rectangular cylinder with inner side lengths a, b; and height h. You also have a solid rectangular cylinder of lengths a, b, the height being irrelevant, and the weight, w. You want to know that given a "perfect" fit, how far down will the weight fall before it comes to rest given the pressure force it has generated.

If this is the case, then we can do it, but we must make some assumptions. Firstly, the gas compression is isentropic; that is, no heat is generated and the process is reversible with no loss in energy. Secondly, the fit is perfect and frictionless; that is, no heat is generated and all of the volume of gas below the mass stays below.

We define the distance the weight has fallen measured from the bottom (just slightly easier, you'll see) as x, and the pressures above and below as $$p_1$$ and $$p_2$$ respectively.

The weight will come to equilibrium when the force on the top equals the force on the bottom. There are three forces, two pressure forces and weight. Our free-body gives us:
$$(p_2 - p_1)a b = w$$
The isentropic compression of a gas is given by
$$\frac{p_2}{p_1} = \left(\frac{V_1}{V_2}\right)^\gamma$$
Where the V are before and after volumes. We solve for the pressure below the piston:
$$p_2 = p_1\left(\frac{V_1}{V_2}\right)^\gamma$$
Since we don't care about a volume fraction, rather we'd know the height, we substitute in before and after volumes.
$$p_2 = p_1\left(\frac{hab}{xab}\right)^\gamma$$
the width and height cancel giving is a simple ratio of heights:
$$p_2 = p_1\left(\frac{h}{x}\right)^\gamma$$
We then plug this value for the pressure below the "piston" back into the force equilibrium equation:
$$\left[ p_1\left(\frac{h}{x}\right)^\gamma - p_1\right] a b = w$$
Now, if I did my algebra correctly, the solution for x would be:
$$x = \frac{h}{ \left[\frac{w}{p_1 a b} + 1\right]^{\frac{1}{\gamma}} }$$
The distance from the top y, could then be simply given by h-x. Also realizing that $$p_1$$ should just be ambient pressure. oops, almost forgot, $$\gamma$$ is the ratio of specific heats, defined as $$c_p / c_v$$

Of course...there could be typos everywhere, but I think the process should be OK.

5. Aug 11, 2010

### ACPower

Well you know the final pressure will be, in your example 25lbs divided by the area (4 square ft). You can also assume the pressure times volume remains constant, so if the area A is 4sq.ft , then A*100ft*(initial pressure) = A*(final height)*Pressure. Cross off those As and: final height = 100ft*(initial pressure)/final pressure.
or:
Hf =Hi*Pi*A/w

Where Hf, Hi are final and initial heights, Pi is the initial (atmospheric) pressure, A is the area and w is the weight.

OOOPS. I forgot to add in the original pressure. The Pf is actually Pi + w/A, not just w/A as above.

Last edited: Aug 12, 2010
6. Sep 16, 2010

### VA80232

Thank You, You have encouraged me to go back and take some math courses. Your formulas