# Energy of a compressed gas

1. Oct 11, 2007

### RKT

Guys ,

Pl. help me out with this one. I got a vessel of vol. V and its filled with air to a pressure P. I wanna know the energy contained in the compressed gas. The gas is gonna be suddenly discharged via an orifice and I wanna know how much work it can do.

I had imagined that the energy would be given by E = P x V however the number arrived by this calc. falls way short of another calc. done by a friend. I estimated the energy contained in a 490 cc vessel filled with air @ 200 bar as 9800 Joules. The friend puts it at abt 24,000 + Joules. He uses the formula for work done during adiabetic expansion.

Now I am wondering ...... did I take only one 'type' of energy into account (pressure) while doing the calc. ? Should heat/thermal energy also be included ? Energy of molecules ? I did not take the heat energy as I figured that any heat generated was basically due to the applied pressure after all. So if I calculated energy using (P x V) it would automatically take that effect into account.

Dash it, this is really confusing. Someone pl. help ........................

2. Oct 11, 2007

### vanesch

Staff Emeritus
The energy of a (perfect) gas under pressure is a tricky thing!
In fact, a perfect gas is such that its energy is only dependent on its temperature U(T) and not on the pressure. But of course, a vessel under pressure can deliver work. So where's the issue ? It depends on how you expand the gas. If you expand it through some diffusion plug, the temperature (of a perfect gas, not of real gasses...) will remain the same, and the energy content of the gas will then remain the same. No work is done. It is btw an irreversible process.
However, if you let it expand adiabatically (reversibly) you've actually constructed a heat machine: the gas expands and COOLS (lowers its energy content), to transform its heat energy stored into mechanical expansion energy. But you can even go further. Let this cold gas, after it expanded, heat up again by thermal contact with the environment. Its temperature will rise, and the pressure will rise too. So you can let it expand again. This time, your (cold) gas is transforming heat energy from the environment to work.
So this energy is not "available in the gas" ; what was actually available was "lack of entropy" in order to drive this conversion of heat energy into work.

3. Oct 12, 2007

### RKT

Here's what I am trying to do. I wanna know the efficiency of the rig. Thats why I need to know the energy before I open the orifice. You must have heard of pneumatic guns, they are quite a craze these days. We are tryin to estimate the efficiency of a particular model.

4. Oct 14, 2007

### Shooting Star

There is not much information here. You have to give at least the volume of the gun barrel. That should be known if you know the length and the cross-section.

One simple approximation would be to treat the pressure P as constant throughout the process, since possibly the volume of the barrel is small compared to the pressure chamber, and the pressure may not change significantly.

Then energy imparted to the pellet would be V*(P – Pa), where Pa is the atmospheric pressure, and V is the volume of the barrel.

5. Oct 14, 2007

### stewartcs

Holding the temperature constant at 60 degrees F, and with 490 CC at 200 bar I came up with approximately 19,237 Joules using NIST data.

6. Oct 14, 2007

### RKT

Stewartcs,

Is that number the energy of the gas 'as it is' compressed in the vessel ? If yes, can you pl. tell me how you calculated it ?

Shootingstar,

I didnt quite understand why you need the barrel vol. I dont need to calculate the energy of the pellet, if thats what you are trying to do. I already know that.
Pl. note : I only need to know if (pressure x volume) will fully account for the energy contained by a compressed gas in a vessel. If no, whats it missing and how do I take that into account as well ?

7. Oct 14, 2007

### Shooting Star

In an ideal gas whose volume is V, the internal energy is (3/2)PV. I really do not know what you are going to do with this formula in the airgun. Only a part of the energy can be converted to mechanical work, depending on the path followed between the initial and the final state. A lot of things matter.

Suppose the pressure ouside is more than the pressure in the compressed gas chamber. Then would you be able to shoot using your compressed air? So, outside pressure is a factor to be taken into. How much the gas will expand and in what way is also important. That's why the volume of the barrel is a factor in the problem, or so I think.

8. Oct 14, 2007

### stewartcs

That is the internal energy in Joules at the specified criteria (200 bar, 60 degrees F, 490 CC).

I personally use a NIST database that gives the internal energy based on specified inputs. The program uses various equations of state to calculate the various data that it is capable of outputting.

One equation of state for finding the internal energy is the Ideal Gas Law, which Shooting star has quoted you already. The 3/2 is however for the average translational kinetic energy of any kind of molecule in an ideal gas. Air is composed mainly of diatomic molecules so we assume it to behave like one. 7/5 or 1.4 is the typical constant to use for an Ideal Gas equation of state for air since it has 5 degrees of freedom. However, I wouldn't use the Ideal Gas Law because it doesn't hold true for every situation. Real gases behave quite differently under higher pressure and varying temperature. That is why your friend said to use an Adiabatic relation to determine the internal energy.

CS

Last edited: Oct 15, 2007
9. Oct 15, 2007

### vanesch

Staff Emeritus
Actually, this can be re-written as 3/2 RT per mole (given that for a mole, PV = RT). So, indeed, for an ideal gas, only temperature matters for the internal energy. But, as said before, the way you convert part of THIS energy to mechanical work, and how you can use it to convert part of OUTSIDE energy to mechanical work, depends on the exact expansion technique.

10. Oct 15, 2007

### RKT

So, lets say that I had a hypothetical airgun that is 100 % efficient in converting the compressed gas energy to useful work. Assume zero losses. From my viewpoint, the only energy that I'll call useful is pellet KE = 0.5 x (pellet mass) x (velocity)^2. Lets also assume that it fires only one shot per fill.

So my hypothetical airgun (at 100 % eff.) should impart a muzzle energy of (3/2)PV to the pellet, is that right ? Something like the concept of the speed of light being the ultimate speed of anything. Can never exceed that. In this case, (3/2)PV is the absolute max. energy that can be imparted to the pellet ?

Pl. disregard the practical aspect of this discussion, in case you're wondering. I know my question is largely abstract. It is imperative that understand what 100 % eff. for a hypothetical model means in the first place, before I think any further.

11. Oct 18, 2007

### Shooting Star

The problem with your picture is that you are trying to think of a box of compressed gas as a compressed spring, which when released will release the entire stored PE as KE. As I have said earlier, work done by a gas depends on both the initial and final states, and how it reaches the final state from its initial state.

You are not able to understand why I asked about the volume of the barrel. Suppose the barrel is very small in length. Then the gas will expand but the pellet will be ejected after travelling only a short distance and would not have the gained much energy from the pressure of the gas. Also, how much energy the gas will release will depend upon the final volume, which in this case is the volume of the compression chamber plus the volume of the barrel. After that, i.e., after the pellet has been ejected, the gas is still expanding but that is not useful energy to you. So, we have to consider what happens up to the point when the gas has fully occupied the barrel.

Since the process is quick, we’ll consider it to be adiabatic. Then, PV^g=K, where I am writing ‘g’ for gamma. For air, g=7/5. The value of K can be obtained by putting in the initial values of P and V, which you had given.

Work done= integral (P-Pa)dV between V1 and V2, where V1 is the volume of the compression chamber and V2 is (V1+volume of barrel), and Pa is the outside pressure.

After integrating, W=K[V2^(-g+1)-V1^(-g+1)]/(-g+1) – Pa(V2-V1).

You know all the values. Convert to proper units. Remember, this is an idealization.

Hope this helps.

12. Oct 22, 2009

### BlueWaterGuy

Does anyone know if the energy in a cylinder of compressed gas/s (air) contains more energy or less energy than was used to compress it. doe is depend on nature of gas being compressed. A cylinder of compressed O2 by itself clearly would have a higher energy content than the energy consumer in compressing it, right. But I don't get the sense, and can't calculate the energy in a cylinder of compressed air, a mixture of gases. Any Ideas or formulas out there.

13. Oct 22, 2009

### stewartcs

The increase in energy will be slightly less than the energy required to compress it do to irreversibilities in the compression process. The Conservation of Energy always applies. The energy is simply transformed to another form from that which is being supplied to compress it.

You cannot create energy by compressing a gas.

Hope this helps.

CS

14. Jan 14, 2012

### alexandrs35

Gents,

I'll add my two cent here as well. In addition i hope somebody will clarify my confusion.

RKT, I am working on that same problem. I'm designing an air gun and trying to design it as efficient as possible.
My advice, read about exergy and try to define how will you measeure the efficiency of your system, i.e. what are you going to compare the KE of the pellet to?
Also read the study of Seigel on light gas hyper velocity gun published in 1965. It is very technical but it answers a lot of question.

Here is what bugs me though...
Take a look at this paper:
http://www.efcf.com/reports/E14.pdf
on page 6 there is a table that summs up the total energy needed to compress 90m^3 of air from Pa to 300bar.

As expected the adiabatic(isentropic) compression consumes the most amount of energy: 277MJ, which later drops to 51MJ as the gas cools and gives off 226MJ to the environment.

Now...if i let the gas expand very quickly, i.e. isentropically i can only get 51MJ out of the system. If i manage to expand it slower the system manages to suck back some of the heat back while still doing work. Thus the efficiency goes up.

Now...take a look at the table summing up the work from expansion on page 9. The max technical work i can get is only 25MJ!?

where did the other 26MJ go?
It is impossible to measure the total internal energy thus we only can account for the energy difference. The system can hold 51MJ at 20C and 300bar and can hold 226MJ more at 20C and 1 atm. But it's really useless to us at that equilibrium stage.

Now...something tells me that the author is still right and I have missed something. What am i missing?

15. Mar 18, 2013

### pyralfugue

Sorry I'm late.

The problem can be approximated well by assuming an ideal gas under adiabatic expansion (depending on the gas being used). CO2 and Air will be fine for this method. There are two problems though, since the firing mechanism is not specified:

If a fixed volume of CO2 is used, the pressure will drop as the projectile is moved down the barrel. Assuming no gas escapes past the projectile, no friction in the barrel, and no loss of pressure from an orifice plate, the moving boundary work should look something like:

W=integral(p dv) where p is a function of both specific volume and temperature by pv=RT. Using the polytropic equations of state for Ideal gasses we can skip some nasty multivariable calculus and get:

W=[mR/(1-n)](T2-T1) where m is the mass of the gas, R is the gas specific constant, and n is given by the polytropic process equation:
T2/T1 = (v2/v1)^(1-n)

T1 is known (hopefully) and T2 must be gotten by looking up reference properties for the specific gas being used. Reference properties are used by:

vRef2=vRef1*(v2/v1) where v1 is the initial specific volume of the gas and v2 is the final specific volume of the gas (dependent on the length of the barrel), where vRef1 is taken at the same temperature as v1 (and naturally vRef2 is the same temp as v2, giving T2).

The work output of the gas under these conditions will equate to the energy transferred to the projectile.

Alternatively, if the orifice is left open from the compressed gas cylinder, then the pressure in the barrel is nominally constant. This would make the calculation much more annoying, as the mass flow rate would need to be accounted for, but fortunately we can then simply use newton's 2nd law where the applied force is pressure * cross sectional area.

I think your biggest accuracy problem will be with fluids problems like orifice flows, which are a little too complex to work without having a lot more info.
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9043775796890259, "perplexity": 647.91350083596}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860127407.71/warc/CC-MAIN-20160428161527-00221-ip-10-239-7-51.ec2.internal.warc.gz"}