How much energy (in Joules) is stored in a standard 12g CO2 cartridge?

[Lupin]

TL;DR Summary

A standard 12 gram cartridge contains both liquid and gaseous CO2 at 850psi. Assuming we are venting to atmosphere at sea level how much energy can be extracted from the cartridge?

A standard 12 gram cartridge contains both liquid and gaseous CO2 at 850psi. Assuming we are venting to atmosphere at sea level, how much energy can be extracted from the cartridge?

We know it will expand to 12 grams * 22.4 liters/ 1 mole (44grams) = 6.1 liters. But how much energy did we extract to get there?

 

[berkeman]

Welcome to PF.

Fun problem. I would do a quick order-of-magnitude calculation with a compressed spring analogy, but you will get much better answers from others here.

BTW, is this question for your schoolwork?

 

[Juanda]

I believe if we treat it as an open system then it's possible to find the extractable energy by realizing what's the change in enthalpy.

I gotta leave now so I can't check if what I recall is right or run the numbers to find what energy is that. Enthalpy for CO2 is tabulated all over the internet so it's just comparing the room temperature conditions at atmospheric pressure with the room temperature conditions at 850PSI.

Maybe someone else will confirm or dismiss what I just posted. I'm definitely not an expert on this.

 

[jbriggs444]

I believe if we treat it as an open system then it's possible to find the extractable energy by realizing what's the change in enthalpy.

For an ideal gas, the energy in the gas is independent of its pressure. It is purely a function of its temperature.

As the ideal gas expands and does [reversible] work on its environment, this requires that it lose temperature.

If you provide thermal energy so that you have an isothermal expansion then you can get energy given by:$$E=P_iV_i \ln \frac{P_i}{P_f} = -P_fV_f \ln \frac{P_f}{P_i}$$where $P_i$ is the initial pressure, $V_i$ is the initial volume, $P_f$ is the final pressure and $V_f$ is the final volume.

You may note that when expanding into a vacuum, the available mechanical energy is infinite. Practical limitations intrude, of course. There is no vacuum good enough. Manufacturing gas-tight but low friction cylinders, pistons and seals that are light years in extent is... challenging.

Also, an isothermal expansion when shooting a pellet using a charge of CO2 is not realistic. You want to calculate using an adiabatic expansion.

 

[Juanda]

@jbriggs444 you propose to solve it by kind of treating it as an expanding closed piston. That's a valid approach but then I believe the mass to be studied should be the one that will remain in the cartridge after the expansion. Paraphrasing Chestermiller:

Using the closed system version of the 1st law of thermodynamics involves recognizing that the final mass of air remaining in the tank has undergone an adiabatic reversible expansion, pushing the air ahead of it (i.e., doing work on it) into the engine, and also recognizing that, during the process, it exchanges no heat with the air ahead of it (since the temperature is uniform in the tank).

That's from this post: https://www.physicsforums.com/threa...-a-pressurised-air-tank.1053707/#post-6910649
In that post, we talked about an adiabatic process but I assume it can be applied to an isothermal process as well.So, first of all, it would be necessary to find the remaining mass in the cartridge. Depending on whether we consider the expansion to be adiabatic or isothermal we will get different answers.

This is the result I get when solving the problem with the closed system approach as you suggested:

CLOSED SYSTEM - ISOTHERMAL PROCESS
$m_0=.00012kg; P_1=5860546Pa; P_2 = P_{atm}=101325Pa; T_1=T_2=298K$
The cartridge volume can be found using the equation state.
$PV=mRT \rightarrow V_1=\frac{mRT_1}{P_1}=1.152*10^{-6}m^3$
After the expansion, the remaining mass $m_r$ will occupy the same volume $V_1=V_2$. It is then possible to find the remaining mass again with the equation of state.
$PV=mRT \rightarrow m_r=\frac{P_2V_2}{RT_2}=2.074*10^{-6}kg$
That remaining mass initially occupied a smaller portion of the cartridge $V_1$. Using the equation of state again I get.
$PV=mRT \rightarrow V_1^*=\frac{m_rRT_1}{P_1}=1.993*10^{-8} m^3$
Finally, work done by the remaining mass can be expressed as you posted.
$W=P_1V_1^* \ln \frac{P_1}{P_2} = -P_2V_2 \ln \frac{P_2}{P_1} = 0.473J$

This is a very small number when compared with the energy obtained by all the mass in the cartridge isothermally expanding from 850PSI to atmospheric pressure but I think it's right. Do you think the method to solve it is correct?

CLOSED SYSTEM - ADIABATIC PROCESS
$m=.00012kg; P_1=5860546; P_2 = P_{atm}=101325Pa; T_1=298K; C_v=657 J/kgK; k=1.289$
Again, it's necessary to find the remaining mass in the cartridge. The cartridge volume is the same as before.
$PV=mRT \rightarrow V_1=\frac{mRT_1}{P_1}=1.152*10^{-6}m^3$
In this case, less information from State 2 is accessible from the beginning. It is then necessary to use equations from the adiabatic expansion.
$v_1 = \frac{V_1}{m}=0.0096 m^3/kg$
$P_1v_1^{k}=P_2v_2^{k} \rightarrow v_2 = e^{\frac{\ln \frac{P_1v_1^k}{P_2}}{k}} =0.224 m^3/kg$
Then, the remaining mass in the cartridge will be:
$m_r=\frac{V_1}{v_1}=5.153*10^{-6}kg$
Also, once $v_2$ is known, it's possible to find $T_2$.
$PV=mRT \rightarrow Pv=RT \rightarrow T_2 = \frac{P_2v_2}{R} = 119.98K$
Finally, the output work will be the same as the change in internal energy since there is no heat involved (adiabatic process).
$W=-m_rC_v\Delta T = 0.602J$

Somehow, I obtained that the output work is greater if the expansion is adiabatic instead of isothermal. That feels wrong in my head. The origin of that here I guess is that the remaining mass in the cartridge is greater in the adiabatic expansion because the cartridge is way cooler so the CO2 is denser.I initially proposed using enthalpy to solve the problem with energy transport and control volumes but while trying to run the numbers I realized I'm not capable of setting the equations to get the same results. Shouldn't both methods be equivalent? I need to work a little longer on that approach to find the breakthrough.

 

[Juanda]

This problem keeps haunting me when I'm in bed. The amount of energy extracted from the gas depends on the extraction process.

Since OP is asking about the upper limit of how much can be extracted I believe @jbriggs444 is right. If the cartridge was hooked to a piston, then all that mass would contribute to the work obtained from expansion. In fact, I recently worked on a similar problem to find the optimal barrel length to have a projectile pushed by an expanding gas as efficiently as possible.
(Related post: Optimization of barrel length in pneumatic cannons)

On the other hand, the work I calculated is the one done by the gas remaining in the cartridge after the expansion as I believe I understood from the explanation given by @Chestermiller (post #5 for details) but I don't see how that translates to extracted energy from the gas. I see it's the remaining gas pushing on the gas ahead but it feels more energy can be extracted from the gas ahead as well. I can't understand how knowing the work done by the remaining gas in the cartridge could be useful.

I guess I'll give it some more time in my head and eventually create a separate post to discuss it if I can't manage to fully understand it by myself. Probably it'll have to wait for the weekend though.

 

[Chestermiller]

This problem keeps haunting me when I'm in bed. The amount of energy extracted from the gas depends on the extraction process.

Since OP is asking about the upper limit of how much can be extracted I believe @jbriggs444 is right. If the cartridge was hooked to a piston, then all that mass would contribute to the work obtained from expansion. In fact, I recently worked on a similar problem to find the optimal barrel length to have a projectile pushed by an expanding gas as efficiently as possible.
(Related post: Optimization of barrel length in pneumatic cannons)

On the other hand, the work I calculated is the one done by the gas remaining in the cartridge after the expansion as I believe I understood from the explanation given by @Chestermiller (post #5 for details) but I don't see how that translates to extracted energy from the gas. I see it's the remaining gas pushing on the gas ahead but it feels more energy can be extracted from the gas ahead as well. I can't understand how knowing the work done by the remaining gas in the cartridge could be useful.

I guess I'll give it some more time in my head and eventually create a separate post to discuss it if I can't manage to fully understand it by myself. Probably it'll have to wait for the weekend though.

The analysis I did was for an open system, and part of the analysis focused on the gas remaining the the vessel at any time during the adiabatic venting. That is not the case here.

For a 12 g CO2 cartridge, the internal volume is only ~16 cc, meaning that, at room temperature, initially, the CO2 is almost all liquid at a specific volume of about 1.3 cc/gm. 850 psi is very close to the critical pressure, and, at that pressure, the saturated volume of the vapor is only about 5.2 cc/gm. So if all the liquid changed to vapor at this pressure, the volume change would only be about 45 cc. So most of the work is going to be in isothermally expanding the superheated vapor from 850 psi to 14.7 psi. I don't think that the ideal gas law is accurate enough for this.

 

[Chestermiller]

To include real gas effects, I would use the principle of corresponding states, with PV=nzRT, with z the compressibility factor. $$dW=PdV=d(PV)-VdP=nRTdz-nRTz\frac{dP}{P}=nRTdz-nRT(z-1)\frac{dP}{P}-nRT\frac{dP}{P}$$
Integrating, we have: $$W=nRT\ln{(P_i/P_f)}+W^R$$where the residual work R (due to non-ideal effect) is given by $$W^R=nRT(z_f-z_i)+nRT\int_{P_i}^{P_f}{(z-1)\frac{dP}{P}}$$

 

[Juanda]

The analysis I did was for an open system, and part of the analysis focused on the gas remaining the the vessel at any time during the adiabatic venting. That is not the case here.

My main itch remains: even if the math were right (work done by the remaining gas in the cartridge) that number is not especially relevant here. Is that right? I need to revisit the post about the pressurized tank and compare them to see how in that case it was important.

For a 12 g CO2 cartridge, the internal volume is only ~16 cc, meaning that, at room temperature, initially, the CO2 is almost all liquid at a specific volume of about 1.3 cc/gm. 850 psi is very close to the critical pressure, and, at that pressure, the saturated volume of the vapor is only about 5.2 cc/gm. So if all the liquid changed to vapor at this pressure, the volume change would only be about 45 cc. So most of the work is going to be in isothermally expanding the superheated vapor from 850 psi to 14.7 psi. I don't think that the ideal gas law is accurate enough for this.

I didn't even think about that. The scenario simply didn't cross my mind. I'll make sure I consider that kind of possibility when talking about pressurized gasses in the future.

To include real gas effects, I would use the principle of corresponding states, with PV=nzRT, with z the compressibility factor. $$dW=PdV=d(PV)-VdP=nRTdz-nRTz\frac{dP}{P}=nRTdz-nRT(z-1)\frac{dP}{P}-nRT\frac{dP}{P}$$
Integrating, we have: $$W=nRT\ln{(P_i/P_f)}+W^R$$where the residual work R (due to non-ideal effect) is given by $$W^R=nRT(z_f-z_i)+nRT\int_{P_i}^{P_f}{(z-1)\frac{dP}{P}}$$

And about this, I trust it's right but just beyond what I can understand at the moment. I'm still focussing on simpler physics when I got some free time and already struggling with it (regarding thermo I'm reading "Thermodynamics An Engineering Approach").
My current job doesn't involve this kind of math so I don't know if I'll get the chance to dive deep enough into thermodynamics to get there. Only time will tell. Still, thanks for taking the time.

 

[Lupin]

I am still struggling with this. For the purpose of the discussion I would ignore the small amount of CO2 left in the 12 gram CO2 powerlet cartridge. (I'm sorry about the English units but that is all I have.) I know I can get about 50 shots out of my Daisy BB gun where a 5.5 grain bb goes 500 ft/sec. That is about 3 ft pounds per shot or, 4.1 Joules. Therefore 50 x 4.1 = 200 joules.
The answer must be much greater than 200 joules because there has to be a lot of loss.
I'm thinking of a spring but I don't have the technical or math knowledge to figure it out. However, I do know how to shoot my Daisy BB gun.
Any simple answer for dummy?
Thanks.

 

[Juanda]

I'd try the approach done by Chestermiller at #8.

To include real gas effects, I would use the principle of corresponding states, with PV=nzRT, with z the compressibility factor. $$dW=PdV=d(PV)-VdP=nRTdz-nRTz\frac{dP}{P}=nRTdz-nRT(z-1)\frac{dP}{P}-nRT\frac{dP}{P}$$
Integrating, we have: $$W=nRT\ln{(P_i/P_f)}+W^R$$where the residual work R (due to non-ideal effect) is given by $$W^R=nRT(z_f-z_i)+nRT\int_{P_i}^{P_f}{(z-1)\frac{dP}{P}}$$

@Chestermiller how can we know the function $z(P)$ in the range $[P_i, P_f]$ for that isothermal process? I think it's the last piece of the puzzle to be able to compute maximum extractable work from that CO2 cartridge in the described expansion process.
$$W^R=nRT(z_f-z_i)+nRT\int_{P_i}^{P_f}{(z-1)\frac{dP}{P}}$$

$n$ can be found from the mass of CO2, its pressure and temperature.
$T$ I guess we're saying it's ambient temperature so 24ºC should work.
$P_i$ is the 850PSI from the loaded cartrigde.
$P_f$ is 1atm after it expands to ambient pressure.
$R$ is just a constant and it's a matter of choosing the right one (I like using $m$ instead of $n$ because mols make no sense to me).

Unless I'm missing something else, once $z(P)$ is known I'd be able to find the value.

 

[Chestermiller]

It follows from the Law of Corresponding states. This says that, for all substances, to a good approximation, z is same function of reduced temperature and reduced pressure, $$z = z (T_r,P_r)$$where $$T_r=\frac{T}{T_c}$$ and $$P=\frac{P}{P_c}$$where the subscript c signifies the critical point of the substance. A more accurate version involves the Pitzer "ascentric" factor $\omega$ which is likewise tabulated for each substance in thermodynamics books: $$z=z(T_r, P_r,\omega)=z^{(0)}(T_r,P_r)+\omega z^{(1)}(T_r,P_r)$$Graphs of $z^{(0)}$ and $z^{(1)}$ are found in most Thermodynamics books such as Smith and Van Ness, Introduction to Chemical Engineering Thermodynamics or online. Many books have graphs of the entire integral as a function of the reduced temperature and pressure.

 

[Juanda]

Thanks. I'll try to work out the results in the coming days.
I'll check my books first to try finding $z = z (T_r,P_r)$. If it's not there, I'll go to the sources you provided.

 

[Chestermiller]

Thanks. I'll try to work out the results in the coming days.
I'll check my books first to try finding $z = z (T_r,P_r)$. If it's not there, I'll go to the sources you provided.

 

[Juanda]

View attachment 331304

I just found one although yours seems to be more complete (it goes further in the reduced pressure and higher in Z). Also, yours has more curves in it compared to what I found in my book.

The only advantage I can see in the one from my book is that there is a grid in the plot.

I think I got an overall idea of how to solve the problem. Choose a curve based on $T_r$ which we're considering constant because $T$ is constant through the expansion and $T_{cr}$ is a property of the gas that's also constant. Then, I'll divide the curve $z$ into two straight lines to integrate. It'd be just an approximation of what the integral would be but it'd provide some rough numbers.

I'll have to leave it for today though. Time is again running on me.

 

[Lupin]

I didn't realize this was going to be so complicated. I'm learning a lot. I figured there would be an easy calculation that was based upon knowing the mass of the CO2: 12g in liquid state; the temperature: assumed to be constant at room temp 24C; and atmospheric pressure: 1 bar. Then there would be some calculation that either started with 12 grams of CO2 at STP, 6.1 liters, and slowly compressed it to liquid state at constant temperature calculating the required energy to do it. Or, starting with liquid CO2 and letting it slowly expand and push a piston until the pressure under the piston is at or close to 1 atmosphere.

Or counting shots from a Daisy BB gun. 200 joules.

 

[Juanda]

I didn't realize this was going to be so complicated. I'm learning a lot. I figured there would be an easy calculation that was based upon knowing the mass of the CO2: 12g in liquid state; the temperature: assumed to be constant at room temp 24C; and atmospheric pressure: 1 bar. Then there would be some calculation that either started with 12 grams of CO2 at STP, 6.1 liters, and slowly compressed it to liquid state at constant temperature calculating the required energy to do it. Or, starting with liquid CO2 and letting it slowly expand and push a piston until the pressure under the piston is at or close to 1 atmosphere.

Or counting shots from a Daisy BB gun. 200 joules.

For the "easy" calculation I'd use the approximation from @jbriggs444 in post #4. An isothermal expansion of an ideal gas.
Regarding my contribution in post #5, I think it's better to just ignore it. I don't think it's wrong but I don't see how it could have some physical meaning here.
Lastly, for a more accurate but complex approach, @Chestermiller provided a method in post #8. Again it's an isothermal expansion but includes real gas effects. Not all the gas will be expanding like that as he said but most of the work will come from that expansion so I'll treat it as if all the gas followed that process. It's a new method for me so I'm interested in seeing the differences between the 2 methods.

Properties of the gas.
$R_{CO_2}=0.1889 \frac{kJ}{kgK}$; $T_{cr}=304.2K$; $P_{cr}=7.39MPa$.

Known information from the problem definition.
$P_i=850PSI + 1atm = 5.961MPa$; $P_f=1atm=0.101MPa$; $T_i=T_f=T=24^{\circ}C=297K$; $m=12g=12*10^{-3}kg$; $V_i=\frac{mRT}{P_i}=0.00011276 m^3$; $V_f=\frac{mRT}{P_f}=0.00663381 m^3$.

ISOTHERMAL EXPANSION OF AN IDEAL GAS
$$W=\int_{i}^{f}PdV=\int_{i}^{f}\frac{mRT}{V}dV=mRT\ln\frac{V_f}{V_i}=mRT\ln\frac{P_i}{P_f}=2738.9J$$

ISOTHERMAL EXPANSION INCLUDING REAL EFFECTS.
First I need to define the $z$ factor for this problem.

Temperature is constant so using $T_R=\frac{T}{T_{cr}}=0.976 \approx 1$ I can choose the lowest curve which is the closest to that value.
Second, I find in which interval of $P_R$ the expansion happens to linearize $z$ with either one or two straight lines. $P_{R_i}=\frac{P_i}{P_{cr}}=0.806$ and $P_{R_f}=\frac{P_f}{P_{cr}}=0.0137$. It's then possible to conclude we're on the left side of that curve and only one straight line will be enough to approximate the problem. I'm eyeballing the values from the graph which is not ideal but I see this mostly as an exercise to learn. Feel free to try more accurate measurements if necessary and do the proper integral without the linearization if necessary.

Now that $z$ is known, we can tackle the expression for work.
$$W=mRT\ln{(P_i/P_f)}+W^R$$
$$W^R=mRT(z_f-z_i)+mRT\int_{P_i}^{P_f}{(z-1)\frac{dP}{P}}$$

For clarity, I'll solve it in chunks. First, this part is the same as the ideal isothermal expansion
$$mRT\ln{(P_i/P_f)}=2738.9J$$

Then, the first component of $W_R$.
$$mRT(z_f-z_i)=262.14J$$

Finally, the last component of $W_R$ which contains the integral. First, I'll express $z$ in terms of $P$ to solve the integral.
$$z=-0.491P_R+0.996= -0.491\frac{P}{P_{cr}}+0.996$$
$$mRT\int_{P_i}^{P_f}{(z-1)\frac{dP}{P}}=mRT\int_{P_i}^{P_f}{(-0.491\frac{P}{P_{cr}}+0.996-1)\frac{dP}{P}}=272.5J$$

That's easily solvable analytically but I cheated to be a little faster and solved it numerically online to avoid writing the steps.
Anyway, assuming the previous calculations are OK, now we have all the components of $W$.
$$W=2738.9+262.14+272.5=3273.54J$$

So it seems it's possible to extract about 20% more energy from the "real expansion" than from the "ideal expansion". Whether that's right or not I can't judge due to my lack of experience in these problems. Maybe I committed an error while inputting the numbers or in the derivation but I couldn't find it while checking the numbers myself.

 

[Lupin]

My goodness! Thank you for boiling the math down to a number I can understand. W = 3273 Joules?!
I am shocked that it is so high. The Daisy BB gun must be so inefficient! My simple test of shooting it until it runs out of power (about 100 shots) yielded a total kinetic energy work of about 200 Joules. Can we really be that far apart?
I figured when the BB exits the barrel there is a lot of CO2 that vents behind it. So a ballpark would be about he same as the kinetic energy of the BB. That brings my 200 J up to 400J. Then there is the inefficiency of the valving. Again, I figure 50% efficiency so the 400J doubles again to 800J stored in the cartridge. .
This is certainly an interesting problem. Maybe it can be used on a physics exam some time.

 

[Juanda]

My goodness! Thank you for boiling the math down to a number I can understand. W = 3273 Joules?!
I am shocked that it is so high. The Daisy BB gun must be so inefficient! My simple test of shooting it until it runs out of power (about 100 shots) yielded a total kinetic energy work of about 200 Joules. Can we really be that far apart?
I figured when the BB exits the barrel there is a lot of CO2 that vents behind it. So a ballpark would be about he same as the kinetic energy of the BB. That brings my 200 J up to 400J. Then there is the inefficiency of the valving. Again, I figure 50% efficiency so the 400J doubles again to 800J stored in the cartridge. .
This is certainly an interesting problem. Maybe it can be used on a physics exam some time.

Don't trust my math just yet. As I said, I see no errors with it but my experience is extremely limited. It is in fact my first time working with an expansion that considers "real effects". Maybe in the following days, someone with more authority will verify what I posted is right or point out the errors.

Regarding the inefficiencies, you might be interested in this other thread I posted a while ago which is very related to what you're dealing with.
Optimization of barrel length in pneumatic cannons

In that mental exercise, I used some simplifications and an adiabatic expansion instead of an isothermal expansion. But it might still help you to understand better what's going on with your BB gun. To be precise, I believe one of the big causes for the inefficiencies must be the length of the barrel is too short. In an ideal world, you'd want the barrel to be just long enough so your gasses in the barrel expand up to atmospheric pressure (ignoring things like friction in the barrel) but that typically results in barrels too long to be manageable.
All I said is completely dependent on the definition of the problem so I'd need to derive a model of this situation to really be able to judge but that's too much for me now. With what's been posted so far, I believe you already got the approximation you were looking for.