Air pressure to energy, am I missing something?

[Theravenhouse]

Hello!
I've come across this forum looking for an answer to an odd question related to energy production. I have no intention of creating infinite energy so don't worry.

Imagine a very strong balloon with a cord wrapped tightly around it. The balloon expands pulling the cord which spins a turbine. All good so far. Now if I want to use the air escaping the balloon to spin a different turbine am I losing energy from the cord setup? As far as I can tell, both are using the psi to generate energy so it feels like they should cancel each other out but I don't see why they would. I assume this is a simple question with an obvious answer so I do apologize but it's been confusing me for several days

 

[BiGyElLoWhAt]

For 1) I'm pretty sure there are losses to heat in the balloon stretching and contracting.
There would also be frictional losses from the air against the balloon, and say you had 2 setups, both the same and pushed more air into 1 balloon, they would eventually come to equilibrium.

 

[Theravenhouse]

For 1) I'm pretty sure there are losses to heat in the balloon stretching and contracting.
There would also be frictional losses from the air against the balloon, and say you had 2 setups, both the same and pushed more air into 1 balloon, they would eventually come to equilibrium.

But what is the cord setup doing to the energy produced by the air escaping? Does it lose anything from the cord?

 

[Lnewqban]

... Imagine a very strong balloon with a cord wrapped tightly around it. The balloon expands pulling the cord which spins a turbine. All good so far. Now if I want to use the air escaping the balloon to spin a different turbine am I losing energy from the cord setup?...

The way I see it:
The compressor that you use to inflate your balloon must provide extra energy to spin turbine #1 (work #1) via the wrapped cord, besides inflating the balloon (work #2).
Deflating the balloon transfers most of latent work #2 (elastic potential energy) into turbine #2.

 

[BiGyElLoWhAt]

The chord setup is either going to be stretching which generates heat or it's going to be sliding around the balloon which has friction, and therefore heat. Heat/friction = energy loss

 

[Theravenhouse]

The way I see it:
The compressor that you use to inflate your balloon must provide extra energy to spin turbine #1 (work #1) via the wrapped cord, besides inflating the balloon (work #2).
Deflating the balloon transfers most of latent work #2 (elastic potential energy) into turbine #2.

Am I essentially just reducing the volume of the balloon and using up that extra psi at an earlier point in the process?

 

[DaveC426913]

Whoa. Back up a second.

The balloon expands pulling the cord

Why does the balloon expand? Is it rising?

Now if I want to use the air escaping the balloon

What? Air is escaping the balloon? So it's not sealed?
OK, so it's not going to expand - it's going to contract.

I think you would do well to think through your setup carefully.

The biggest thig to examine that that a balloon does not generate energy. you might be able get some potential energy out of a system (say, a rising balloon pulling a string) but:
1] you had to put that energy in by doing work - eg. filling the balloon
2] it's not a cycle. It'll work once and then stop.

 

[Theravenhouse]

Whoa. Back up a second.

Why does the balloon expand? Is it rising?What? Air is escaping the balloon? So it's not sealed?
OK, so it's not going to expand - it's going to contract.

I think you would do well to think through your setup carefully.

The biggest thig to examine that that a balloon does not generate energy. you might be able get some potential energy out of a system (say, a rising balloon pulling a string) but:
1] you had to put that energy in by doing work - eg. filling the balloon
2] it's not a cycle. It'll work once and then stop.

I'll clarify a lot:
This idea is for utilizing energy from already pressurized gas. It is not intended to create net energy but rather offset the energy used in pressurizing gas. Take propane for example, it has a volume 270 times that of it's container. So releasing it from a pressurized container into a balloon fills it up just fine.

That is when the balloon expands, it is not rising in my idea. It expands with compressed gas.

The balloon would have to be resealable (likely something more complex than a party balloon), so it could fill up then be expelled through a turbine then return to being sealed for future use.

 

[hutchphd]

both are using the psi to generate energy so it feels like they should cancel each other out

What do you mean? Describe this in detail and we can start a discussion.
Waving your hands rapidly just moves the hot air.
Work done by pressure always involves a volume change

 

[Dale]

Now if I want to use the air escaping the balloon to spin a different turbine am I losing energy from the cord setup?

No. But of course once the air starts turning the first turbine you will only be getting energy at the second turbine.

By the way, the first turbine is confusing people because turbines are usually turned by gas and not strings. Your setup would be more understandable if it were described as follows:

A pressurized gas is injected into a piston and work is extracted from the piston as it expands. When it is expanded it is still under pressure. The pressure is released and used to drive a turbine.

 

[Theravenhouse]

No. But of course once the air starts turning the first turbine you will only be getting energy at the second turbine.

By the way, the first turbine is confusing people because turbines are usually turned by gas and not strings. Your setup would be more understandable if it were described as follows:

A pressurized gas is injected into a piston and work is extracted from the piston as it expands. When it is expanded it is still under pressure. The pressure is released and used to drive a turbine.

Thank you! That is a much better way to explain it.

 

[Dale]

Thank you! That is a much better way to explain it.

No problem! FYI, such a setup is perfectly fine. It does not violate any physical laws, although it probably would be discarded for engineering reasons.

 

[sophiecentaur]

The balloon expands pulling the cord which spins a turbine. All good so far.

"All good", as long as you say what it is that is making the balloon expand. You have to give a reason / mechanism for any change to a system, if you want to discuss what happens to the Energy.
You may have a 'reason' in your head but you need to be explicit.
You could, for instance, just be slackening the cord, allowing the balloon to expand (releasing some Potential (stored) Energy in the compressed air or you could be pumping air into the balloon / heating it up / changing the air pressure etc. etc.. The sign of the Energy flow is important.
Physics doesn't allow for Sky Hooks or Free Energy

 

[256bits]

A pressurized gas is injected into a piston and work is extracted from the piston as it expands. When it is expanded it is still under pressure. The pressure is released and used to drive a turbine.

Thank you! That is a much better way to explain it.

A double expansion engine.

The Titanic had a triple expansion steam cycle, and/or four when the turbine was in use.
http://www.titanicology.com/Titanica/TitanicsPrimeMover.htm

A way to get as much out of the shoveling of tons of coal needed to power the vessel.

 

[sophiecentaur]

The Titanic had a triple expansion steam cycle

As do the steam turbines used in Power Stations, I believe, and aircraft turbo fans etc etc.

 

[Stavros Kiri]

Hello!
I've come across this forum looking for an answer to an odd question related to energy production. I have no intention of creating infinite energy so don't worry.

Good! Welcome, thanks (that's why the forum is here for) and good that you're not ambitious for that infinite energy (etc.)

Imagine a very strong balloon with a cord wrapped tightly around it. The balloon expands pulling the cord which spins a turbine. All good so far. Now if I want to use the air escaping the balloon to spin a different turbine am I losing energy from the cord setup? As far as I can tell, both are using the psi to generate energy so it feels like they should cancel each other out but I don't see why they would. I assume this is a simple question with an obvious answer so I do apologize but it's been confusing me for several days

Indeed strange set-up (and I would be curious enough to see what your original motivation was etc. or how exactly your original idea came about etc. – although, I think you answered some of these [and more, etc.] in a subsequent post of your's that I liked ...), but the answer, AS YOU SAY, is INDEED really simple! :

These energies do not of course cancel each other out but they just add up, giving you the total energy. So you're not actually losing energy (in total), but yes you're losing it from one part (i.e. the cord set-up, in this case), coming out of the total energy ...

So it's actually like in the 'Conservation of Energy' arguments etc. ... , the reason basically of your not being able to create that infinite energy (or energy out of nothing) that you said yourself ...
[So bassically you had the question answered yourself ...]
Cheers

P.S. + energy possibly lost e.g. in friction and heat (etc.), as already mentioned by other member etc. ...

 

[sophiecentaur]

Comment from a pedantic old gimmer: If you think you've found some inconsistency in this 'Energy Thing' then you have to assume that you've forgotten something in the equation you have in your head.
Apart from in smart Alec mode, when e=mc² gets brought into things, you can rely on Energy (i.e. all of it) is conserved. Also you can forget Relativity until your understanding of the classical stuff is pretty damn good. So just make a list of everything and eliminate a few that don't matter in your model and the change will be Zero. Sometimes it can be a simple matter of a sign you have used.

 

[Daemius]

Hello!
I've come across this forum looking for an answer to an odd question related to energy production. I have no intention of creating infinite energy so don't worry.

Imagine a very strong balloon with a cord wrapped tightly around it. The balloon expands pulling the cord which spins a turbine. All good so far. Now if I want to use the air escaping the balloon to spin a different turbine am I losing energy from the cord setup? As far as I can tell, both are using the psi to generate energy so it feels like they should cancel each other out but I don't see why they would. I assume this is a simple question with an obvious answer so I do apologize but it's been confusing me for several days

If the external pressure doesn't reduce (from rising higher in the atmosphere for example) you need to pump air inside the baloon (consume energy) for it to expand and therefore produce energy by pulling the cord. By releasing the air inside the ballon, it will start deflating, no more producing air from the cord. You could then use the airflow to generate energy until the baloon empties out. The total energy produced by the cord and the airflow cannot exceed the energy required to inflate the baloon. Take also in account that the more you compress air (the air in the baloon will be under pressure), the higher its temperature rises, creating a temperature difference with the outside and higher rates of heat dissipitation (thermal energy loss).

 

[Dale]

If the external pressure doesn't reduce (from rising higher in the atmosphere for example) you need to pump air inside the baloon (consume energy) for it to expand and therefore produce energy by pulling the cord. By releasing the air inside the ballon, it will start deflating, no more producing air from the cord. You could then use the airflow to generate energy until the baloon empties out. The total energy produced by the cord and the airflow cannot exceed the energy required to inflate the baloon. Take also in account that the more you compress air (the air in the baloon will be under pressure), the higher its temperature rises, creating a temperature difference with the outside and higher rates of heat dissipitation (thermal energy loss).

Good first post @Daemius
Welcome to PF!