# Homework Help: Air compression question

1. May 23, 2013

### Thomas38740

1. The problem statement, all variables and given/known data
How far will a piston fall down a vertical cylinder 100 feet in length? If the cylinder is 10" in diameter, the piston is 100lbs, the cylinder has a bottom and is capable of fluid isolation, filled with ambient air at 80 degrees F, at sea level, the cylinder has friction free slide-ability.

2. Relevant equations

3. The attempt at a solution
I have look in FAQ. I have tried to solve the question.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
I have had an education. I am a senior citizen veteran who does not always understand the new ways. I have not used my education in years. I am trying to keep my mind young as possible.

2. May 23, 2013

### Basic_Physics

The piston will actually oscillate up and down in the cylinder, so you might be interested in the lowest point it will reach? The reason for this being that the air can be compressed like a spring.

3. May 23, 2013

### Thomas38740

Thank You

I am interested in where it finds average equilibrium approximately. Example 10 feet down, 25 feet down, 80 feet down the cylinder.

4. May 23, 2013

### Basic_Physics

It seems to me you are interested in the point where the pressure applied by weight of the piston and the ambient pressure on its top surface will be equal to the pressure exerted by the air inside of the cylinder.

If so you can use Boyle's law which states that the pressure, p, and volume, V, of a confined gas is inversely proportional to each other at constant temperature or

p1 V1 = p2 V2

The 1st state is then when the gas in the cylinder is not compressed by the piston and the 2nd state when the pressure balances the pressure of the weight of the piston and the atmospehric pressure on top of it.

Last edited: May 23, 2013
5. May 23, 2013

### Thomas38740

Thank You.

Can this formula be worked out so I can see how it is done and see also the answer. If I am taught how to fish once I should be able to do it again and again on my own.

6. May 23, 2013

### Thomas38740

the volume I think would be BxH=V or 3.14x5inches squared=78.5
The height is 100feet or 1200 inchesx78.5=94,200sq. inches
Ok know I have the volume. The pressure at the top is 14.7 ambient pressure.
So P1 =14.7 psi and volume or V1 = 94,200 sq inches
so 14.7x94200= 1,384,740

How does this tell me how far the piston goes down the cylinder?

7. May 23, 2013

### haruspex

A good habit to get into is to keep everything in symbols, only plugging in numbers at the end. Makes it easier to spot mistakes and for others to comprehend your work.
Cancellation may also mean there's less calculation to do. In the present case, P1 V1 = P2 V2 can be expanded to P1 H1 A = P2 H2 A, A being the constant cross-sectional area, which then cancels to produce P1 H1 = P2 H2.
You have P1 and H1, so you just need P2. What will P2 be when in balance with the weight of the piston and the atmosphere above it?

8. May 23, 2013

### Thomas38740

Thank You

I think I can figure this out. I am going to try tomorrow with a fresh mind. You seem to have done a good job of explaining it. I just need to rest and try it again tomorrow. In the mean time I am going to join this forum and support it. I do a lot of walking and thinking

9. May 24, 2013

### Basic_Physics

If state 1 is just as the piston is inserted into the cylinder we have that

pa V1

where pa is the atmospheric air pressure and V1 is the total volume of the cylinder which we can calculate with

V1 = A h1

where A is the area of the circular top or bottom. In this case h1 = 100 ft. We now have that

pa V1 = pa A h1

For state 2 when the pressure exerted by the piston is balanced by the pressure in the cylinder we have that

(pa + pp) A h2

where pp is the pressure exerted by the weight of the piston and h2 is the required height. The pressure exerted by the piston can be evaluated via

pp = 4 x 100 lbs / ∏ D2

Where the area of the piston is given by ∏ D2/4 in total we then have that

pa A h1 = (pa + 4 x 100 lbs / ∏ D2) A h2

so that the circular areas cancel out

pa h1 = (pa + 4 x 100 lbs / ∏ D2) h2

pa 100 ft = (pa + 4 x 100 lbs / ∏ 100 inch2) h2

pa 100 ft = (pa + 4/∏ lbs/inch2) h2

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