Undergrad Adiabatic expansion work far exceeds isobaric of same volume, why?

[MysticDream]

If the piston is massless and frictionless, then by Newton's 2nd law, the force per unit area exerted by the external atmosphere on the outside face of the piston $P_{ext}$ must be equal in magnitude to the force per unit area exerted by the gas on the inside face of the piston. That means that the work done by the internal gas on the inside face of the piston must be $$dW=P_{ext}dV$$

Could you give me the formula for the net work done on the piston in the adiabatic case?

If I'm not mistaken, you seem to be in agreement with jack action. I'd just like to clarify.

 

[Chestermiller]

Could you give me the formula for the net work done on the piston in the adiabatic case?

If I'm not mistaken, you seem to be in agreement with jack action. I'd just like to clarify.

In the adiabatic reversible case, the gas within the cylinder is only slightly removed from thermodynamic equilibrium along the entire process path, so the force per unit area on the inside face of the piston is described in this situation by the ideal gas law: $$dW=P_{ext}dV=\frac{nRT}{V}dV$$Note that this means that the external pressure $P_{ext}$ must be forced to change as a function of volume in the adiabatic reversible case.

 

[MysticDream]

In the adiabatic reversible case, the gas within the cylinder is only slightly removed from thermodynamic equilibrium along the entire process path, so the force per unit area on the inside face of the piston is described in this situation by the ideal gas law: $$dW=P_{ext}dV=\frac{nRT}{V}dV$$Note that this means that the external pressure $P_{ext}$ must be forced to change as a function of volume in the adiabatic reversible case.

Thanks.

So the net work done on the piston only (not the atmosphere) for the adiabatic process, would be:

$$W=(P_{1}*V_{1}-P_{2}*V_{2})/(7/5-1)-P_{ext}dV$$

 

[Chestermiller]

Thanks.

So the net work done on the piston only (not the atmosphere) for the adiabatic process, would be:

If it is a massless frictionless piston, in this adiabatic reversible version of the process, the net work done on the piston is zero. $$Work\ done\ by\ gas\ in\ cylinder\ on\ piston=\frac{(P_1V_1-P_2V_2)}{\gamma-1}$$
$$Work\ done\ by\ externally\ controlled\ pressure\ on\ piston = -\int{P_{ext}(V)dV}=-\frac{(P_1V_1-P_2V_2)}{\gamma-1}$$For a massless frictionless piston, the net work done on the piston is always zero, throughout the process.

 

[MysticDream]

If it is a massless frictionless piston, in this adiabatic reversible version of the process, the net work done on the piston is zero. $$Work\ done\ by\ gas\ in\ cylinder\ on\ piston=\frac{(P_1V_1-P_2V_2)}{\gamma-1}$$
$$Work\ done\ by\ externally\ controlled\ pressure\ on\ piston = -\int{P_{ext}(V)dV}=-\frac{(P_1V_1-P_2V_2)}{\gamma-1}$$For a massless frictionless piston, the net work done on the piston is always zero, throughout the process.

Ah yes, but if the piston had any mass at all, that formula would hold true, correct?

 

[Chestermiller]

Ah yes, but if the piston had any mass at all, that formula would hold true, correct?

If it were an adiabatic reversible process, even if the piston had mass, if the cylinder were horizontal, the net work would be zero both at the end of the process (at final thermodynamic equilibrium) and during the process.

For an irreversible expansion, the net work done on a piston with mass during the process (again assuming horizontal cylinder) would be $$W(V)=\int_{V_1}^V{\frac{F_g}{A}dV}-P_{ext}(V-V_1)=\frac{1}{2}mv^2$$assuming that the externally applied pressure is constant. However, if the system has reached final thermodynamic equilibrium at the end of the process, such that the kinetic energy of the piston has been damped by the gas, then we would be left with $$W(V_2)=\int_{V_1}^{V_2}{\frac{F_g}{A}dV}-P_{ext}(V_2-V_1)=0$$and $$\int_{V_1}^{V_2}{\frac{F_g}{A}dV}=P_{ext}(V_2-V_1)$$

 

[MysticDream]

If it were an adiabatic reversible process, even if the piston had mass, if the cylinder were horizontal, the net work would be zero both at the end of the process (at final thermodynamic equilibrium) and during the process.

For an irreversible expansion, the net work done on a piston with mass during the process (again assuming horizontal cylinder) would be $$W(V)=\int_{V_1}^V{\frac{F_g}{A}dV}-P_{ext}(V-V_1)=\frac{1}{2}mv^2$$assuming that the externally applied pressure is constant. However, if the system has reached final thermodynamic equilibrium at the end of the process, such that the kinetic energy of the piston has been damped by the gas, then we would be left with $$W(V_2)=\int_{V_1}^{V_2}{\frac{F_g}{A}dV}-P_{ext}(V_2-V_1)=0$$and $$\int_{V_1}^{V_2}{\frac{F_g}{A}dV}=P_{ext}(V_2-V_1)$$

I’m lost now. How is work on the piston calculated then? Perhaps I used the wrong term. By net work, I meant work done on the piston.

If the piston had mass and the external environment was a complete vacuum (0 psia) then work would be done on the piston to move the piston since it has inertia.

Logically, if the external pressure was 14.7 psia, there would still have to be work done on the piston mass, but work would also be done on the external environment.

 

[MysticDream]

Are you referring to if I let the piston go beyond equal pressure on both sides? I could see in that case that the kinetic energy on the piston could lower the pressure internally relative to the external pressure. I mean to calculate only the work done on the piston up until the pressures are equal on both sides.

 

[Chestermiller]

I’m lost now. How is work on the piston calculated then? Perhaps I used the wrong term. By net work, I meant work done on the piston.

By net work, do you mean work done by the gas in the cylinder on the piston, or co you mean work done by the gas in the cylinder minus work done by the outside atmosphere. Which of these "net works" do you think determines the final state of the gas?

If the piston had mass and the external environment was a complete vacuum (0 psia) then work would be done on the piston to move the piston since it has inertia.

Logically, if the external pressure was 14.7 psia, there would still have to be work done on the piston mass, but work would also be done on the external environment.

Do you think that, for a piston with mass, the piston would still be moving at very long times? Or would it exhibit damped oscillation about the final equilibrium volume in approaching final steady state (until its KE is zero)?

 

[MysticDream]

By net work I meant work done by the gas in the cylinder on the piston. This is what I’m trying to calculate.

I only intended to calculate the work done on the piston up until the final volume I had originally specified. There would be no reversal of the piston and no oscillations. I understand this would happen if a piston was released and left to continue its motion, however, that is not what I’m trying to calculate.

 

[Chestermiller]

By net work I meant work done by the gas in the cylinder on the piston. This is what I’m trying to calculate.

OK. Correct. This is the work done by the gas on its surroundings.

I only intended to calculate the work done on the piston up until the final volume I had originally specified. There would be no reversal of the piston and no oscillations. I understand this would happen if a piston was released and left to continue its motion, however, that is not what I’m trying to calculate.

At the time the piston reaches the specified original final volume and is stopped, the gas will not be in thermodynamic equilibrium. You understand that additional adjustment takes place within the gas (at constant volume) until the gas reaches its final thermodynamic equilibrium state. During this readjustment, the pressure of the gas will eventually deviate from the outside gas pressure, and the temperature will also change. So, even though you have stopped the piston (and the work done), you have not stopped the process (until final thermodynamic equilibrium is reached).

For now, let's temporarily confine attention to the case of a massless piston. Let's consider the case of a piston with mass later.

 

[MysticDream]

OK. Correct. This is the work done by the gas on its surroundings.

At the time the piston reaches the specified original final volume and is stopped, the gas will not be in thermodynamic equilibrium. You understand that additional adjustment takes place within the gas (at constant volume) until the gas reaches its final thermodynamic equilibrium state. During this readjustment, the pressure of the gas will eventually deviate from the outside gas pressure, and the temperature will also change. So, even though you have stopped the piston (and the work done), you have not stopped the process (until final thermodynamic equilibrium is reached).

For now, let's temporarily confine attention to the case of a massless piston. Let's consider the case of a piston with mass later.

Yes, I understand what you are saying about additional adjustment.

I'm not sure if you mean the "piston only" by the phrase "work done by the gas on its surroundings" or if you mean the "piston and the atmosphere". Technically both are the surroundings. Again, I'm intending to understand how to calculate only the work done on the piston. In other words, the "useful work" that could be used to drive something mechanically. If this is what you mean by net work, then we meant the same thing.

 

[Chestermiller]

Yes, I understand what you are saying about additional adjustment.

I'm not sure if you mean the "piston only" by the phrase "work done by the gas on its surroundings" or if you mean the "piston and the atmosphere". Technically both are the surroundings. Again, I'm intending to understand how to calculate only the work done on the piston. In other words, the "useful work" that could be used to drive something mechanically. If this is what you mean by net work, then we meant the same thing.

I mean the force exerted by the gas on the piston integrated over the displacement of the piston. This work is given by the equation $$W=\int_{V_1}^{V_2}{\frac{F_g}{A}dV}=P_{ext}(V_2-V_1)$$where, in your system $$P_{ext}=14.7\ psia$$

 

[MysticDream]

I mean the force exerted by the gas on the piston integrated over the displacement of the piston. This work is given by the equation $$W=\int_{V_1}^{V_2}{\frac{F_g}{A}dV}=P_{ext}(V_2-V_1)$$where, in your system $$P_{ext}=14.7\ psia$$

If a piston has mass, is there a difference between the work done on the piston and the work done on the atmosphere? What is the useful work that could be transferred to a mechanical device like a pump? In the adiabatic case, it seems to me that the only useful work that could drive a mechanical system would be the work done on the piston mass. The work done on the atmosphere is wasted. Therefore the useful work is the Total work done by the gas minus the portion of the work done on the atmosphere.

 

[MysticDream]

I noticed above you said for now let’s focus on a massless piston, and we can consider the case of a piston with mass later. Perhaps we should specify exactly which case we are referring to when describing it.

I understand in the case of a massless piston, that the total work done by the gas is also done on the external atmosphere.

In the case of a piston with mass, the work done by the gas is partly on the piston mass (because it has inertia) and partly on the external atmosphere, if I’m correct.

 

[Chestermiller]

If a piston has mass, is there a difference between the work done on the piston and the work done on the atmosphere? What is the useful work that could be transferred to a mechanical device like a pump? In the adiabatic case, it seems to me that the only useful work that could drive a mechanical system would be the work done on the piston mass. The work done on the atmosphere is wasted. Therefore the useful work is the Total work done by the gas minus the portion of the work done on the atmosphere.

I don't know what "useful work" means or how the concept applies in this context. In this system, the piston will have to come to a stop one way or another.

 

[Chestermiller]

I noticed above you said for now let’s focus on a massless piston, and we can consider the case of a piston with mass later. Perhaps we should specify exactly which case we are referring to when describing it.

I understand in the case of a massless piston, that the total work done by the gas is also done on the external atmosphere.

In the case of a piston with mass, the work done by the gas is partly on the piston mass (because it has inertia) and partly on the external atmosphere, if I’m correct.

In this system, does the piston eventually come to a stop or doesn't it? It is either stopped by, say, a small annular protrusion into the cylinder at V2, or by being damped to a stop by viscous forces in the gas. Either way, at final steady state, the piston mass is no longer moving and its final kinetic energy is zero.

 

[MysticDream]

In this system, does the piston eventually come to a stop or doesn't it? It is either stopped by, say, a small annular protrusion into the cylinder at V2, or by being damped to a stop by viscous forces in the gas. Either way, at final steady state, the piston mass is no longer moving and its final kinetic energy is zero.

It seems we have gotten far away from my original question in the OP. I want to know how much work (in the adiabatic case) was done on the piston (with mass) by the gas up until the point at which it is at the final volume. The external pressure is atmospheric pressure and constant. Whether the piston stops or not does not matter. If the piston was released horizontally it would fly a certain distance. If it was stopped it would hit against a stop and deform. Either way, the piston has gained kinetic energy up until that point at the final volume. This is my only question.

 

[Chestermiller]

It seems we have gotten far away from my original question in the OP. I want to know how much work (in the adiabatic case) was done on the piston (with mass) by the gas up until the point at which it is at the final volume. The external pressure is atmospheric pressure and constant. Whether the piston stops or not does not matter. If the piston was released horizontally it would fly a certain distance. If it was stopped it would hit against a stop and deform. Either way, the piston has gained kinetic energy up until that point at the final volume. This is my only question.

Like I said, let's do it without the piston having mass first, OK? If we can't do it for this case, we will not be able to do it for the more complicated case of a piston with mass.

You want to do it for the case in which the piston is stopped once the final volume matches that for the adiabatic reversible case, right?

 

[MysticDream]

Like I said, let's do it without the piston having mass first, OK? If we can't do it for this case, we will not be able to do it for the more complicated case of a piston with mass.

You want to do it for the case in which the piston is stopped once the final volume matches that for the adiabatic reversible case, right?

Yes

 

[Chestermiller]

Yes

OK. For this adiabatic irreversible process, the first law if thermodynamics gives us $$\Delta U=nC_v(T_f-T_1)=-W$$where $$W=\int_{V_1}^{V_2}{\frac{F_g}{A}dV}=P_{ext}(V_2-V_1)=(14.7)(44.9)(0.113)=74.6\ J$$ and $$n=\frac{P_1V_1}{RT_1}$$Combining these equation, we obtain:$$\frac{P_1V_1}{(\gamma-1)}\left[\frac{T_f}{T_1}-1\right]=-W$$Solving this equation for the ratio of the final temperature to the initial temperature then gives: $$\frac{T_f}{T_1}=1-\frac{W(\gamma-1)}{P_1V_1}=1-\frac{(76.4)(0.4)}{(24.7)(100)(0.113)}=0.8905$$

OK so far?

 

[jack action]

I'm going to chime in again as I feel both of you have difficulties understanding each other. I hope I won't add to the confusion.

For the OP, the piston of cross-sectional area $A$ is connected to something that offers a resistive force $F$ - like a crankshaft for example. The crankcase pressure would be at a steady $P_{atm}$. On the other side of the piston, there is a combustion chamber with pressure $P$ whose volume varies from $V_1$ to $V_2$. In one case, $P$ follows an adiabatic expansion, in the other, it follows an isobaric expansion. So the energy equation would be:
$$\int_{V_1}^{V_2}{PdV}=P_{atm}(V_2-V_1) + F\frac{(V_2-V_1)}{A}$$
(For simplicity, I'm assuming both sides of the piston have the same cross-sectional area.)

The "net work" the OP refers to is $F\frac{(V_2-V_1)}{A}$, i.e. the useful work going to the crankshaft, where $\frac{(V_2-V_1)}{A}$ is the piston displacement.

If the piston has a mass, piston acceleration must also be considered, if any.

Feel free to tell me if I don't understand either of you correctly.

 

[Chestermiller]

I'm going to chime in again as I feel both of you have difficulties understanding each other. I hope I won't add to the confusion.

For the OP, the piston of cross-sectional area $A$ is connected to something that offers a resistive force $F$ - like a crankshaft for example. The crankcase pressure would be at a steady $P_{atm}$. On the other side of the piston, there is a combustion chamber with pressure $P$ whose volume varies from $V_1$ to $V_2$. In one case, $P$ follows an adiabatic expansion, in the other, it follows an isobaric expansion. So the energy equation would be:
$$\int_{V_1}^{V_2}{PdV}=P_{atm}(V_2-V_1) + F\frac{(V_2-V_1)}{A}$$
(For simplicity, I'm assuming both sides of the piston have the same cross-sectional area.)

The "net work" the OP refers to is $F\frac{(V_2-V_1)}{A}$, i.e. the useful work going to the crankshaft, where $\frac{(V_2-V_1)}{A}$ is the piston displacement.

If the piston has a mass, piston acceleration must also be considered, if any.

Feel free to tell me if I don't understand either of you correctly.

You don’t understand what I have been saying correctly. Before you posted this, I thought the OP and I had reached consensus. I hope what you have posted here does not introduce new confusion.

 

[MysticDream]

OK. For this adiabatic irreversible process, the first law if thermodynamics gives us $$\Delta U=nC_v(T_f-T_1)=-W$$where $$W=\int_{V_1}^{V_2}{\frac{F_g}{A}dV}=P_{ext}(V_2-V_1)=(14.7)(44.9)(0.113)=74.6\ J$$ and $$n=\frac{P_1V_1}{RT_1}$$Combining these equation, we obtain:$$\frac{P_1V_1}{(\gamma-1)}\left[\frac{T_f}{T_1}-1\right]=-W$$Solving this equation for the ratio of the final temperature to the initial temperature then gives: $$\frac{T_f}{T_1}=1-\frac{W(\gamma-1)}{P_1V_1}=1-\frac{(76.4)(0.4)}{(24.7)(100)(0.113)}=0.8905$$

OK so far?

Yes.

 

[MysticDream]

I'm going to chime in again as I feel both of you have difficulties understanding each other. I hope I won't add to the confusion.

For the OP, the piston of cross-sectional area $A$ is connected to something that offers a resistive force $F$ - like a crankshaft for example. The crankcase pressure would be at a steady $P_{atm}$. On the other side of the piston, there is a combustion chamber with pressure $P$ whose volume varies from $V_1$ to $V_2$. In one case, $P$ follows an adiabatic expansion, in the other, it follows an isobaric expansion. So the energy equation would be:
$$\int_{V_1}^{V_2}{PdV}=P_{atm}(V_2-V_1) + F\frac{(V_2-V_1)}{A}$$
(For simplicity, I'm assuming both sides of the piston have the same cross-sectional area.)

The "net work" the OP refers to is $F\frac{(V_2-V_1)}{A}$, i.e. the useful work going to the crankshaft, where $\frac{(V_2-V_1)}{A}$ is the piston displacement.

If the piston has a mass, piston acceleration must also be considered, if any.

Feel free to tell me if I don't understand either of you correctly.

Yes, I was referring to the useful work that could be transferred to the crankshaft.

I thought your formula that subtracted the isobaric “contraction” of the external atmosphere from the total work of the gas (using the adiabatic formula) is what would give me the work transferred to a crankshaft. I thought chestermiller would be in agreement with that. If he has further insight to offer, I’m all ears.

 

[Chestermiller]

Yes.

Well, the next step is to calculate the final equilibrium pressure within the cylinder using the ideal gas law: $$\frac{P_fV_2}{T_f}=\frac{P_1V_1}{T_1}$$So, $$P_f=P_1\frac{V_1}{V_2}\frac{T_f}{T_1}=(24.7)\frac{100}{144.9}(0.8905)=15.2\ psia=0.5\ psig$$So by stopping the piston motion before reaching its natural equilibrium position, we obtain a final pressure higher than the pressure against whcihthe gas expanded, 14.7 psia. So, in this case, the final temperature and final pressure are higher than in the adiabatic reversible case.

 

[Chestermiller]

Yes, I was referring to the useful work that could be transferred to the crankshaft.

I thought your formula that subtracted the isobaric “contraction” of the external atmosphere from the total work of the gas (using the adiabatic formula) is what would give me the work transferred to a crankshaft. I thought chestermiller would be in agreement with that. If he has further insight to offer, I’m all ears.

The most fundamental concept of good modeling practice is to start simple, and add complexity later as needed. Why? Well first of all, if you are unable to solve a simple version of the problem, you certainly won't be able to solve a more complex version. Plus, once you solve the simple version, you will have some understanding developed and some quantitative results under your belt in a short period of time and with minimal effect. These results can then provide a basis of comparison as more complexity is added. This is called the KISS principle: Keep It Simple, Stupid.

 

[MysticDream]

Well, the next step is to calculate the final equilibrium pressure within the cylinder using the ideal gas law: $$\frac{P_fV_2}{T_f}=\frac{P_1V_1}{T_1}$$So, $$P_f=P_1\frac{V_1}{V_2}\frac{T_f}{T_1}=(24.7)\frac{100}{144.9}(0.8905)=15.2\ psia=0.5\ psig$$So by stopping the piston motion before reaching its natural equilibrium position, we obtain a final pressure higher than the pressure against whcihthe gas expanded, 14.7 psia. So, in this case, the final temperature and final pressure are higher than in the adiabatic reversible case.

Yes, I understand.

 

[jack action]

You don’t understand what I have been saying correctly. Before you posted this, I thought the OP and I had reached consensus. I hope what you have posted here does not introduce new confusion.

I'm sorry to get involved but this thread is very confusing and my understanding is that you are not explaining the process described in the OP. At first, you seemed to be describing this:

For the adiabatic irreversible isobaric expansion you were considering in the second case, I believe you intended to start out at 10 psi gauge again, but in this case to suddenly drop the external pressure to zero psi gauge and to hold that external pressure constant until the gas had expanded by the same volume as in the first case. So $\Delta V=44.9\ in^3=0.000736\ m^3$ and $P_{ext}=14.7\ psi=101.325\ kPa$, and the work is W = 74.6 J.

The OP answered back to this with:

Sorry for the misunderstanding, but that is not the first calculation I was intending to do. I was expanding a 10 psig volume of gas adiabatically against an external pressure of 0 psig. In both cases the external pressure was 0 psig.

This could be visualized as a cylinder with a piston in it. Both cylinders and pistons are of the same exact size, will travel the same volume while doing the work, and have the same starting pressure and external pressure. The only difference is that one cylinder will expand adiabatically to 0 psig, and the other will maintain pressure.

You say the piston goes down while the pressure is at 0 psig; I believe the OP is saying the piston goes down while the pressure is maintained at 10 psig.

When the OP talks about "external pressure", he means the pressure on the other side of the piston.

I believe the isobaric case in the OP is NOT adiabatic.

Then you came back with:

If the piston is massless and frictionless, then by Newton's 2nd law, the force per unit area exerted by the external atmosphere on the outside face of the piston $P_{ext}$ must be equal in magnitude to the force per unit area exerted by the gas on the inside face of the piston. That means that the work done by the internal gas on the inside face of the piston must be $$dW=P_{ext}dV$$

Here you seem to say that the piston is set free, just being pushed around by the different pressures in the chambers on each side of the piston. I believe the OP expects that one side produces more work than the other takes, the resultant energy being transferred to the piston doing useful work by pushing something else.

My understanding is that the piston is not free. Its motion is restricted by some external mechanism.

The work done by the internal gas on the inside face of the piston is expected to be greater than the work done by the external atmosphere on the outside face of the piston, not equal.

Finally, here you say:

You want to do it for the case in which the piston is stopped once the final volume matches that for the adiabatic reversible case, right?

To which the OP answered "Yes". The case to be compared is supposed to be isobaric by definition. How come you end up in post #56 with a process where the final pressure is different from the initial pressure? To me, both the initial and final pressures are 24.7 psia. That is the constraint of the problem.

If the OP is not confused, I am.

 

[MysticDream]

If the OP is not confused, I am.

I think he is intending to explain everything by starting with basics rather than just answering my initial question, which is fine, although I was confused at first. I believe you answered my initial question already. My misunderstanding was that I thought the adiabatic formula I was using described the useful work that could be done on a piston, but did not realize that work was also being done (unfortunately wasted) on the external atmosphere. By subtracting that from the total work done by the gas, I end up with the useful work that would be converted into kinetic energy (on the piston).