1. Sep 29, 2008

### Steve Stone

Hi every one,

When I compress air in a simple bicycle pump (with the outlet sealed) the handle or piston bounces back when i let it go.

The energy released by the compressed air inside the pump seems to be less than the energy required to compress it.

Am I right in my observation?

Also, if correct, is there a point of compression where the energy released by the compressed air is greater than the energy required to compress it?

I ask this because it seems that if i compress the air quickly; the piston inside the pump bounces back faster and further than if the air is compressed slowly.

Is this because the air gets warm and expands slightly due to a faster compression?

Also, my physics master tells me that if air is compressed sufficiently then it will combust in the same way as Diesel fuel.

Is this true?

Thanks, Steve

2. Sep 29, 2008

### DeShark

I think the reason it's not bouncing back so far is due to escaped gas rather than any energy intake. The work done in compressing the pump will cause an increase in the internal energy of the gas.

In a closed system, this energy can only be released to the universe by loss of heat, through conduction. There isn't a sufficient amount of time to do this in either of your two processes, so generally the pump would move back to its original position. However, there is another way of releasing this energy and that's through a physical movement of gas out of the bicycle pump (i.e. the seal isn't perfect). More gas would be lost in a greater amount of time being compressed, explaining why the pump doesn't return as much during a slow compression. This seems like a more realistic scenario to me and is my best guess as to why the pump doesn't return completely. Hope this helps...

3. Sep 29, 2008

### jablonsky27

The hand;e/piston wont bounce back all the way up because even after you let go of it, its weight will still be acting on the air. So its still very very slightly compressed.

This certainly isnt true - air doesnt combust. Perhaps he was talkinga bout a diesel engine and you misunderstood him. In a diesel engine, air is compressed(to about 1/8 its original volume). This increases the temperature of the air, and diesel is sprayed onto it. Due to the high temperature, diesel combusts immediately.

4. Sep 29, 2008

### DeShark

I don't think this is relevant. It's only relevant if jablonsky27 decided midway through compression to rotate the pump from horizontal to vertical. Which wouldn't be a fair test.

5. Sep 29, 2008

### Andy Resnick

What you say makes sense- there are going to be viscous (friction) and thermal losses present, so some of the elastic energy is dissipated and lost.

By the way, who is your physics master? Can I get one of those? :)

6. Sep 29, 2008

### stewartcs

Yes. The small difference (loss) is due to irreversibilities in the piston-cylinder system (such as seal friction). Therefore, the amount of energy that is available to push the handle back up is slightly less (in this case) than what you put into it (the other energy was not lost but rather transformed into another form so energy is still conserved).

Energy cannot be created or destroyed, therefore the maximum amount of energy in the air is equal to the total energy it had to begin with (before you compressed it) plus what ever you added to it (transferred by work in this case).

Air is composed (mostly) of Nitrogen and Oxygen, neither which are fuel sources. Without a fuel source, no combustion will occur. He was incorrect to say that.

CS

7. Sep 30, 2008

### Steve Stone

Thanks everyone,

I didn't believe that air would burn. If it did and someone was to put a perverbial match to it; i think we would all be in big trouble!

OK, so if for example air is compressed quickly in a closed system (with no leaks); would the fact that the air gets warm and expands mean that more energy is contained in the air than if it was compressed slowly and didn't get warm?

Thanks, Steve

8. Sep 30, 2008

### DeShark

It's important to understand where the energy is coming from in this situation; that is, who/what is doing work. It's all about thermodynamics. The first law of which states that the change in internal energy equals the work done on the system plus the heat flow into the system (the system being, in your case, the gas and bicylcle pump). You do work when you compress the air. This increases the internal energy of the system. This increase in internal energy is seen as an increase in temperature and pressure. If the walls of the pump are not good insulators and the process is done very slowly, the process is called "isothermal". That is, there is no change in the temperature of the system (because heat is being conducted out to the universe). If the walls are very good insulators (perfect insulators to be precise), the process is known as adiabatic. No heat is lost. Aditionally, if the gas is a so-called "perfect-gas", the volume and temperature in an adiabatic process are related by:

$$VT^\alpha = \operatorname{constant}$$

where alpha is the number of degrees of freedom divided by 2. (degrees of freedom of the gas particles: 3 for a monatomic gas [3 translational, i.e. xyz]; 5 for a diatomic molecule [3 translational, xyz, plus 2 rotational, theta, phi])

So finally, to answer your question: Even if the gas is compressed slowly, it will get warm if the walls are good insulators. However, if the walls are bad insulators, then internal energy will escape in the form of heat. So, yes there will be less energy contained if it's compressed slowly enough that the temperature remains constant.

9. Sep 30, 2008

### Steve Stone

Thanks DeShark.

Im trying to understand how the Stirling hot air engine works and wondering if i could apply the principle to a rotary motion for my project

After what you have said i feel a little more confident.

Thanks, Steve