Adiabatic or Isothermal? Compressed air energy storage

In summary, the conversation revolves around a high school physics project where the student is trying to determine if their setup is adiabatic or isothermal in order to calculate the work done. They have an empty water bottle with a small hole on the bottom and a valve on the top for compressed air. The student is also trying to account for air humidity in their calculations and is looking for a citable source to clarify this. They are specifically interested in the process of the bottle emptying while accelerating a toy car. There is also a discussion about the most inexact part of the calculation being the estimation of the discharge rate of gas through the hole.
  • #1
AEGIS
12
0
Hi,

I'm doing a high school physics project and am trying to figure out if a certain setup that I'm using is adiabatic or isothermal, in order to determine what equation I can use to calculate the work that my setup does-- the threads I've come across so far only explain the difference, but not how to tell whether a certain setup is one or the other.

Setup: I have an ordinary empty water bottle, which has a small hole on its bottom made by a heated paperclip. This hole is small enough such that a thumb tack can fit tightly through it as a 'seal' of sorts. Additionally, the bottle has a valve on its top, through which compressed air can be sent and contained with very marginal loss over relatively short periods of time such as a few minutes or less (which is what I'm dealing with for my setup).

I pump a certain pressure of compressed air into the bottle (say, 30 psi). I then attach it to a toy car and remove the thumb tack, and a stream of pressurized air is expelled, propelling the car. This experiment is performed for an ordinary water bottle (0.5 L), a 1 L bottle, and a 2 L bottle that typically carries pop.

I understand the definitions of 'isothermal' and 'adiabatic', but I still don't understand how to tell which (if either) my project is? This is (unfortunately) far above anything I have or likely will be covering in my high school physics class.

Isothermal indicates that the temperature of the gas in the system is constant throughout.
Adiabatic indicates that there is no heat transfer between the fluid (air, in this case), and the surroundings, which may or may not be isentropic if the process is internally reversible.

If someone could clarify this for me, along with a citable source such as a university or formal research paper (Wikipedia doesn't count), that would be much appreciated. Thank you.
 
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  • #2
Also! I may have to account for air humidity in my calculations, as I will have to calculate the density of the air as per the project's requirements. I found something here (see 'pressure dependence') assuming an isothermal process, but it seemed to apply to an air-water system, which my setup most likely is not. Does this apply, assuming the process is in fact isothermal? If not, are there other assumptions underlying this? I had trouble relations on air humidity based on pressure apart from that of the atmosphere or the aforementioned setup.

Once again, if possible, a citable source would be much appreciated. Thank you.
 
  • #3
Which process are you interested in, (a) filling the bottle, or (b) bottle emptying while accelerating the car?

Regarding correcting for humidity in determining the density: it is not significant, because there is so little water vapor in the air at room temperature.
 
  • #4
Chestermiller said:
Which process are you interested in, (a) filling the bottle, or (b) bottle emptying while accelerating the car?

Regarding correcting for humidity in determining the density: it is not significant, because there is so little water vapor in the air at room temperature.

Definitely the bottle emptying; I'm trying to relate the loss of pressure (flow rate) to the overall acceleration of the car at specific time increments using F = ma, then compare a rough calculation of the 'theoretical' under ideal conditions to the experimental values that I actually obtain. This is obviously a bit more complicated by the fact that the forces acting on the car (force of thrust from the stream of gas and the force of friction) and the mass are both changing, due to the loss of pressurized air from the bottle. I recognize that my calculations won't be exactly the same as my experimental values obtained due to the force of friction etc. being difficult to calculate exactly, but my teacher specifically noted that the size of the % error doesn't matter as long as the process used to arrive at the values is sound and we demonstrate that we've learned how to apply our physics & laboratory concepts with regards to error analysis.

About the density of the air: assuming correcting for humidity is insignificant, does this mean I can use a variation of ideal gas law to determine density? (I found the equation density = pressure/(R*temperatuere).) Additionally, do you have any idea where I might be able to find a citable academic source that states such a correction is insignificant?
 
  • #5
AEGIS said:
Definitely the bottle emptying; I'm trying to relate the loss of pressure (flow rate) to the overall acceleration of the car at specific time increments using F = ma, then compare a rough calculation of the 'theoretical' under ideal conditions to the experimental values that I actually obtain. This is obviously a bit more complicated by the fact that the forces acting on the car (force of thrust from the stream of gas and the force of friction) and the mass are both changing, due to the loss of pressurized air from the bottle. I recognize that my calculations won't be exactly the same as my experimental values obtained due to the force of friction etc. being difficult to calculate exactly, but my teacher specifically noted that the size of the % error doesn't matter as long as the process used to arrive at the values is sound and we demonstrate that we've learned how to apply our physics & laboratory concepts with regards to error analysis.

The most inexact part of your calculation is going to be that of estimating the discharge rate of gas through the hole. This is going to determine the force that the gas exerts on the bottle/car. You need to focus on that. You have an open hole discharging from a container in which there is a gas under pressure.
About the density of the air: assuming correcting for humidity is insignificant, does this mean I can use a variation of ideal gas law to determine density? (I found the equation density = pressure/(R*temperatuere).) Additionally, do you have any idea where I might be able to find a citable academic source that states such a correction is insignificant?
Actually, the equation you gave is for the molar density. The actual mass density is equal to the molar mass of air (29 g) times the molar density. If you are worried about water vapor, look up the equilibrium vapor pressure of water at room temperature. If you multiply this by the relative humidity of the air (say 70%), you get the pressure of the water vapor in the air. You can compare this with atmospheric pressure to decide whether you think the water vapor is significant.
 
  • #6
Chestermiller said:
The most inexact part of your calculation is going to be that of estimating the discharge rate of gas through the hole. This is going to determine the force that the gas exerts on the bottle/car. You need to focus on that. You have an open hole discharging from a container in which there is a gas under pressure.

Okay. Would emptying the bottle be isothermal or adiabatic? I'm uncertain as to whether the above principles apply (i.e. temperature remains constant throughout process, or no heat is lost during process). Depending on which process it is the equations that I can use also differ, it seems.

Chestermiller said:
Actually, the equation you gave is for the molar density. The actual mass density is equal to the molar mass of air (29 g) times the molar density. If you are worried about water vapor, look up the equilibrium vapor pressure of water at room temperature. If you multiply this by the relative humidity of the air (say 70%), you get the pressure of the water vapor in the air. You can compare this with atmospheric pressure to decide whether you think the water vapor is significant.

Does this mean that the density of the air in the bottle is changing as the air is released due to changing pressure? Is this going to be a problem in calculating values for force of thrust, force of friction, mass, etc., some of which are dependent on the air's density or pressure, and will I have to factor this change into any equations I used that depend on these values?
 
  • #7
AEGIS said:
Okay. Would emptying the bottle be isothermal or adiabatic? I'm uncertain as to whether the above principles apply (i.e. temperature remains constant throughout process, or no heat is lost during process). Depending on which process it is the equations that I can use also differ, it seems.
I think it is going to be closer to adiabatic than isothermal, because the time interval is going to be pretty short, so not much heat can enter. But, do it both ways, and see how the results compare.
Does this mean that the density of the air in the bottle is changing as the air is released due to changing pressure?
Sure.
Is this going to be a problem in calculating values for force of thrust, force of friction, mass, etc., some of which are dependent on the air's density or pressure, and will I have to factor this change into any equations I used that depend on these values?
It will have to be taken into account. Whether this is a problem or not depends on your perspective.
 

1. What is adiabatic compressed air energy storage (CAES)?

Adiabatic CAES is a type of energy storage system where compressed air is stored in an underground reservoir, typically a cavern or aquifer, and released to generate electricity. The compression and expansion of air is done without any heat exchange with the environment, making it an adiabatic process.

2. How does adiabatic CAES differ from isothermal CAES?

The main difference between adiabatic and isothermal CAES is the use of heat exchange. In adiabatic CAES, there is no heat exchange during compression and expansion, while in isothermal CAES, heat is exchanged to maintain a constant temperature. This makes adiabatic CAES more efficient as it requires less energy for compression and expansion.

3. What are the advantages of adiabatic CAES?

One of the main advantages of adiabatic CAES is its high efficiency. As there is no heat exchange, the energy used for compression can be almost entirely recovered during expansion, resulting in a high round-trip efficiency. Adiabatic CAES also has a high energy storage capacity and can be easily integrated with existing power grids.

4. Are there any drawbacks to adiabatic CAES?

One potential drawback of adiabatic CAES is the need for a suitable underground reservoir, which can be costly to construct. There may also be concerns about the environmental impact of drilling and creating the reservoir. Additionally, adiabatic CAES may not be suitable for areas with high population density due to safety concerns.

5. How is adiabatic CAES being used in renewable energy systems?

Adiabatic CAES is being used in renewable energy systems as a way to store excess energy generated from renewable sources, such as wind and solar, for use during times of high demand. This helps to balance out the intermittency of renewable energy sources and can help to reduce the need for traditional fossil fuel power plants. Adiabatic CAES can also be used to store excess energy from off-peak times for use during peak demand periods.

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