Question about Flow between Parallel Plates

[Red_CCF]

No. The interface/external pressure always balances for a massless frictionless piston. The variation in external pressure does not have to be applied in discrete steps. This is just a specific example that your book presented. The external force variation can also be applied continuously with time, as long as it is done very slowly. Do you want to try modeling some other applied force variations with time to see how that plays out?

I thought quasistatic process is the step-by-step process we solved for but at infintiessimal step-sizes. Does increasing pressure in dP incrment count as a discrete steps? From my recollection of first year calculus, continuous function y(x) is one where infinitessimal dx variation causes infinitessimal dy change.

With regards to the step about "waiting until equilibrium is reached," I was thinking that although interface/external pressure always equal, pressure inside the cylinder does not instantaneously equal the interface pressure. As we approach quasistatic process, I imagine that the time for cylinder pressure to equal interface pressure to approach dt; is this correct?

Out of curiosity, what other kinds of applied force variation can also work for quasistatic processes?

You can apply any external force variation with time you desire. In the case of the spring-damper model, you already calculated the motion of the piston necessary to hold the force constant. You are asking an experimental question, rather than a conceptual question. Think about putting a force transducer on the piston and controlling its motion so that the measured force varies in the way that you desire. This can be done automatically (with motors and feedback loops) or, in concept, it can be done manually.

Chet

How would one control the motion of a massless/frictionless piston (making it start/stop), if such an object theoretically exists?

Thank you very much

 

[Chestermiller]

I thought quasistatic process is the step-by-step process we solved for but at infintiessimal step-sizes.

No. In a quasistatic process, at each and every point (in time) along the process path from the initial equilibrium state to the final equilibrium state, the system is only slightly removed from being at thermodynamic equilibrium. The process does not have to be carried out in discrete steps.

Does increasing pressure in dP incrment count as a discrete steps? From my recollection of first year calculus, continuous function y(x) is one where infinitessimal dx variation causes infinitessimal dy change.

Certainly, the single large step process is not continuous, according to this definition. In any event, all the discrete step sequences that we analyzed were not continuous.

With regards to the step about "waiting until equilibrium is reached," I was thinking that although interface/external pressure always equal, pressure inside the cylinder does not instantaneously equal the interface pressure.

We already discussed in previous posts that, in irreversible discrete step external force variation processes, the pressure within the cylinder is not uniform spatially and, in addition, viscous dissipation is occurring. In our spring-damper model, the viscous stresses are supporting most of the load change initially, while, at later times, the external force is being supported by the homogeneous gas pressure. In the real world where we are compressing an actual gas, the gas inertia (mass) also plays a role. In earlier posts, we talked about spring-damper-mass analogs that mimic more closely the behavior of an actual gas, but we never got around to analyzing them (and I didn't feel that they would add that much to your understanding).

As we approach quasistatic process, I imagine that the time for cylinder pressure to equal interface pressure to approach dt; is this correct?

If the external force is continuous, all that is required is that the external force varies slowly enough with time.

Out of curiosity, what other kinds of applied force variation can also work for quasistatic processes?

Here are some examples:
$$P_{ext}=P_{Ei}+(P_{Ef}-P_{Ei})\frac{t}{τ}$$
$$P_{ext}=P_{Ei}+(P_{Ef}-P_{Ei})(1-e^{-\frac{t}{τ}})$$
where τ is the characteristic time scale for the process. To approach quasistatic behavior, τ is very large. Try substituting either of these into our model equations for the spring-damper problem, and see what you get for the total work, the quasistatic work, and the work to overcome viscous stresses. You will see that you will come to the same conclusions from these force variations that you did when analyzing a sequence of discrete steps.

How would one control the motion of a massless/frictionless piston (making it start/stop), if such an object theoretically exists?

You can control the motion of any piston (even pistons with mass) if you use devices, such as powerful motors, which have enough oomph.

Chet

 

[Red_CCF]

No. In a quasistatic process, at each and every point (in time) along the process path from the initial equilibrium state to the final equilibrium state, the system is only slightly removed from being at thermodynamic equilibrium. The process does not have to be carried out in discrete steps.

Certainly, the single large step process is not continuous, according to this definition. In any event, all the discrete step sequences that we analyzed were not continuous.

The way quasistatic processes was taught to me from digging through my notes from awhile back is similar to what is in the attached diagram (it neglects viscous effects). In Case 1 if M is pushed onto the piston from its platform, a work of Mzg (or lost energy) is needed to bring the piston back to the height of the platform so the mass can be removed. In Case 2 (two platforms at different heights), only a work on the order of about Mzg/2 is needed since half the mass is removed at lower than initial height. So in general about Mzg/n of "extra" work/irreversibility is needed to bring the piston back to its original height and as n -> ∞ (n number of platforms and discrete masses that add up to M), we approach a reversible/quasistatic process.

The above is overly simplistic but my understanding of quasistatic process is one where the number of steps in the process -> ∞ and per step, ΔP -> dP so I visualize it as an infinite number of discrete steps with infinitessimal step sizes where the sequence of events for each step is similar to that of of the process at finite steps size. Is this simply a different way to looking at it or is it incorrect?

You can control the motion of any piston (even pistons with mass) if you use devices, such as powerful motors, which have enough oomph.

Chet

What is oomph?

Thank you very much

 

[Chestermiller]

The way quasistatic processes was taught to me from digging through my notes from awhile back is similar to what is in the attached diagram (it neglects viscous effects). In Case 1 if M is pushed onto the piston from its platform, a work of Mzg (or lost energy) is needed to bring the piston back to the height of the platform so the mass can be removed. In Case 2 (two platforms at different heights), only a work on the order of about Mzg/2 is needed since half the mass is removed at lower than initial height. So in general about Mzg/n of "extra" work/irreversibility is needed to bring the piston back to its original height and as n -> ∞ (n number of platforms and discrete masses that add up to M), we approach a reversible/quasistatic process.

The above is overly simplistic but my understanding of quasistatic process is one where the number of steps in the process -> ∞ and per step, ΔP -> dP so I visualize it as an infinite number of discrete steps with infinitessimal step sizes where the sequence of events for each step is similar to that of of the process at finite steps size. Is this simply a different way to looking at it or is it incorrect?

There is nothing wrong with envisioning a Quasistitic process as the limit of a sequence of discrete steps as the number of steps becomes infinite. But this is not the only way of approaching this limit. In first-year calculus, we learned about using integration to get the area under a curve by taking the limit of a discrete set of rectangles. But we could also have used a set of trapazoids that have ordinates that are piecewise linear and continuous. Using a continuous force variation for P_ext(t) is somewhat analogous to this. I still contend that it would be of value of you to re-solve the problem with either of the two functionalities that I suggested in my previous post. You will then see how this works.

Incidentally, your interpretation of what is depicted in the figure you attached is incorrect. If you like, we can redo our problem for the spring-damper system in which we add masses to the piston as depicted in your figure. Then you will see how this really should be interpreted. I would also point out that, although the analysis related to the figure did not mention viscous stresses (for obvious reasons, since the students at that level were unaware at that point of what viscous stresses are), the viscous stresses in this situation are not negligible (although it is not necessary to explicitly take them into account). However, they are definitely responsible for preventing the added mass and the gas from oscillating forever, and they are responsible for the added work that the masses must do to compress the gas, over and above the quasistatic work.

What is oomph?

Oomph is slang for vigor or vigorousness or ability to apply as much force we want in any way we wish.

Chet

 

[Red_CCF]

There is nothing wrong with envisioning a Quasistitic process as the limit of a sequence of discrete steps as the number of steps becomes infinite. But this is not the only way of approaching this limit. In first-year calculus, we learned about using integration to get the area under a curve by taking the limit of a discrete set of rectangles. But we could also have used a set of trapazoids that have ordinates that are piecewise linear and continuous. Using a continuous force variation for P_ext(t) is somewhat analogous to this. I still contend that it would be of value of you to re-solve the problem with either of the two functionalities that I suggested in my previous post. You will then see how this works.

Is the main difference in quasistatic process with a continuous force variation P_ext(t) is that it is a direction function of time unlike a sequence of discrete steps? In the case of continuous forces, what is considered analogous to a "step"?

Incidentally, your interpretation of what is depicted in the figure you attached is incorrect. If you like, we can redo our problem for the spring-damper system in which we add masses to the piston as depicted in your figure. Then you will see how this really should be interpreted. I would also point out that, although the analysis related to the figure did not mention viscous stresses (for obvious reasons, since the students at that level were unaware at that point of what viscous stresses are), the viscous stresses in this situation are not negligible (although it is not necessary to explicitly take them into account). However, they are definitely responsible for preventing the added mass and the gas from oscillating forever, and they are responsible for the added work that the masses must do to compress the gas, over and above the quasistatic work.
Chet

What were some of my mis-interpretations? Chances are I forgot about something since I lost my original diagram.

Thank you very much

 

[Chestermiller]

Is the main difference in quasistatic process with a continuous force variation P_ext(t) is that it is a direction function of time unlike a sequence of discrete steps? In the case of continuous forces, what is considered analogous to a "step"?

A quasistatic process does not imply a sequence of discrete steps. It just means that the process is carried out very slowly. What you are missing is that, even in the case of a sequence of discrete steps, there is an underlying time variation that you need to consider. Each time a discrete step is applied, the system needs a certain amount of time to equilibrate. In our spring damper problem, this time interval is on the order of C/k, where C is the damper constant and k is the spring constant. Only the effective time interval over which the deformation is occurring (i.e., the gas is compressing) is relevant. If the sequence involves n steps, the total amount of time to go from the initial equilibrium state to the final equilibrium state is on the order of τ=nC/k. So, the more steps you have, the longer it will take to go from the initial state to the final state, and the slower the effective rate of deformation. In the case of continuous forces, the work performed and the viscous work dissipation need to be calculated over the same total deformation time τ. Under these circumstances, the continuous force variation and the discrete force application are directly comparable. I'm begging you to redo the calculations for the total work and for the viscous work loss, using the continuous force variations that I wrote down a couple of posts ago. We will then be able to compare this with the discrete sequence and see how they directly compare.

What were some of my mis-interpretations? Chances are I forgot about something since I lost my original diagram.

I contend that the analyses we did in previous posts for the case of a piston with mass can help you to work out the correct interpretation without my help.

Chet

 

[Red_CCF]

A quasistatic process does not imply a sequence of discrete steps. It just means that the process is carried out very slowly. What you are missing is that, even in the case of a sequence of discrete steps, there is an underlying time variation that you need to consider. Each time a discrete step is applied, the system needs a certain amount of time to equilibrate. In our spring damper problem, this time interval is on the order of C/k, where C is the damper constant and k is the spring constant. Only the effective time interval over which the deformation is occurring (i.e., the gas is compressing) is relevant. If the sequence involves n steps, the total amount of time to go from the initial equilibrium state to the final equilibrium state is on the order of τ=nC/k. So, the more steps you have, the longer it will take to go from the initial state to the final state, and the slower the effective rate of deformation. In the case of continuous forces, the work performed and the viscous work dissipation need to be calculated over the same total deformation time τ. Under these circumstances, the continuous force variation and the discrete force application are directly comparable. I'm begging you to redo the calculations for the total work and for the viscous work loss, using the continuous force variations that I wrote down a couple of posts ago. We will then be able to compare this with the discrete sequence and see how they directly compare.
Chet

I got myself mixed up a bit due to the time variable. From the original ODE
$$AP_{ext}(t)=AP_{ext}(0)-C\frac{dx}{dt}-kx$$
for
$$P_{ext}=P_{Ei}+(P_{Ef}-P_{Ei})\frac{t}{τ}$$
substitution of the second into the first equation gives
$$(P_{Ei}-P_{Ef})\frac{t}{τ}=C\frac{dx}{dt}+kx$$
The notation is a bit weird as essentially AP_Ei=P_Ei but the P's in the last equation should be total pressure force.
$$x(t)=\frac{P_{Ei}-P_{Ef}}{τ}[\frac{t}{k}-\frac{C}{k^2}+\frac{C}{k^2}e^{\frac{-kt}{C}}]$$
and
$$\frac{dx}{dt}=\frac{P_{Ei}-P_{Ef}}{kτ}[1-e^{\frac{-kt}{C}}]$$
If I multiply both sides of the ODE by dx/dt as we did previously to get work and integrate from t from 0 to infinity the equation blows up due to the linear t variable. So far I haven't been able to spot an arithmatic mistakes so I am probably applying the equation improperly. Can you point me in the right direction?

Thank you very much

 

[Chestermiller]

I got myself mixed up a bit due to the time variable. From the original ODE
$$AP_{ext}(t)=AP_{ext}(0)-C\frac{dx}{dt}-kx$$
for
$$P_{ext}=P_{Ei}+(P_{Ef}-P_{Ei})\frac{t}{τ}$$
substitution of the second into the first equation gives
$$(P_{Ei}-P_{Ef})\frac{t}{τ}=C\frac{dx}{dt}+kx$$
The notation is a bit weird as essentially AP_Ei=P_Ei but the P's in the last equation should be total pressure force.
$$x(t)=\frac{P_{Ei}-P_{Ef}}{τ}[\frac{t}{k}-\frac{C}{k^2}+\frac{C}{k^2}e^{\frac{-kt}{C}}]$$
and
$$\frac{dx}{dt}=\frac{P_{Ei}-P_{Ef}}{kτ}[1-e^{\frac{-kt}{C}}]$$
If I multiply both sides of the ODE by dx/dt as we did previously to get work and integrate from t from 0 to infinity the equation blows up due to the linear t variable. So far I haven't been able to spot an arithmatic mistakes so I am probably applying the equation improperly. Can you point me in the right direction?

Thank you very much

Yes. Don't forget that, for all t ≥ τ, P_ext(t)=P_Ef. So you have to split the solution into two regions. That affects x(t) and dx/dt. (Maybe it would have been easier (mathematically) to use the other P_ext variation I suggested, in which P_ext(t) is changing at all times from t =0 to t = ∞.)

Chet

 

[Chestermiller]

I solved this problem for the case of a non-discrete external force variation given by:
$$P_{ext}(t)=P_{Ef}-(P_{Ef}-P_{Ei})e^{-\frac{t}{τ}}$$
Please see if you can reproduce my results.

For x(t), I got:
$$x(t)=-\frac{(P_{Ef}-P_{Ei})A}{k}\left[(1-e^{-\frac{t}{τ}})-\frac{(e^{-\frac{t}{τ}}-e^{-\frac{kt}{C}})}{[1-\frac{C}{kτ}]}\right]$$
From this, for the total work I got:
$$-W=\frac{(P_{Ef}^2-P_{Ei}^2)A}{2k}+\frac{(P_{Ef}-P_{Ei})^2A} {2k[1+\frac{kτ}{C}]}$$
The first term on the right hand side represents the reversible (quasistatic) work. The second term represents the extra work required to deform the dissipative damper. Note that, as the characteristic time for the process τ→∞, the second term approaches zero.

Note the similarity between these results, and the results we got in posts # 124 and #125 where we considered a discrete sequence of pressure variations. Is it clearer to you now that, to achieve a quasistatic (reversible) deformation, you don't need to use discrete steps?

Chet

 

[Red_CCF]

I solved this problem for the case of a non-discrete external force variation given by:
$$P_{ext}(t)=P_{Ef}-(P_{Ef}-P_{Ei})e^{-\frac{t}{τ}}$$
Please see if you can reproduce my results.

For x(t), I got:
$$x(t)=-\frac{(P_{Ef}-P_{Ei})A}{k}\left[(1-e^{-\frac{t}{τ}})-\frac{(e^{-\frac{t}{τ}}-e^{-\frac{kt}{C}})}{[1-\frac{C}{kτ}]}\right]$$
From this, for the total work I got:
$$-W=\frac{(P_{Ef}^2-P_{Ei}^2)A}{2k}+\frac{(P_{Ef}-P_{Ei})^2A} {2k[1+\frac{kτ}{C}]}$$
The first term on the right hand side represents the reversible (quasistatic) work. The second term represents the extra work required to deform the dissipative damper. Note that, as the characteristic time for the process τ→∞, the second term approaches zero.

Note the similarity between these results, and the results we got in posts # 124 and #125 where we considered a discrete sequence of pressure variations. Is it clearer to you now that, to achieve a quasistatic (reversible) deformation, you don't need to use discrete steps?

Chet

I was able to get the x(t) equation but unable to get the form for the work equation. For the work equation, are we solving:
$$-W=\int{C(\frac{dx}{dt})^2dt}+\int{kx\frac{dx}{dt}dt}$$
where we integrate t from 0 to infinity? If so I probably made some arithmatic errors during expanding/simplification.

With regards to the work equation, is the area A supposed to be squared?

In discrete step, we integrated each step from t = 0 to infinity, essentially saying that each step takes infinite time. As we take the number of steps n to approach infinity, are we basically saying that the process will take ∞² to complete?

What is τ physically? I see that it is synonymous with the number of steps (when pressure is increased in discrete steps), but how does one measure/control this variable?

Thank you very much

 

[Chestermiller]

I was able to get the x(t) equation but unable to get the form for the work equation. For the work equation, are we solving:
$$-W=\int{C(\frac{dx}{dt})^2dt}+\int{kx\frac{dx}{dt}dt}$$
where we integrate t from 0 to infinity? If so I probably made some arithmatic errors during expanding/simplification.

Well, it's easier to integrate the following:
$$-W=A\int_0^∞{P_{ext}(t)\frac{dx}{dt}dt}$$
You can get additional simplification by integrating by parts:
$$P_{ext}(t)\frac{dx}{dt}dt=d(xP_{ext})-x\frac{dP_{ext}}{dt}dt$$

With regards to the work equation, is the area A supposed to be squared?

Ooops. You're right. I'm missing a factor of A. Thanks.

In discrete step, we integrated each step from t = 0 to infinity, essentially saying that each step takes infinite time. As we take the number of steps n to approach infinity, are we basically saying that the process will take ∞² to complete?

In Post #146, I discussed the fact that the "effective" amount of time to complete a single discrete step is C/k. Please read over my post carefully. If you wait until t = 4C/k, the displacement will be 98% of the displacement you get at infinite time. It won't matter significantly if you start the next discrete step then, or wait an infinite amount of time. So the nominal time for a sequence of n discrete steps will be nC/k (or 4nC/k if you prefer).

What is τ physically?

τ is a constant we are using to parameterize how fast we vary the external pressure. A small value of τ means we are varying it very rapidly, and a large value of τ means we are varying it slowly. Physically, τ is the amount of time it takes for P_ext to rise 63% of the way from P_Ei to P_Ef.

I see that it is synonymous with the number of steps (when pressure is increased in discrete steps), but how does one measure/control this variable?

As I said in several previous posts, we can control the applied external pressure variation P_ext(t) to be any function of time we desire.

Chet

 

[Red_CCF]

In Post #146, I discussed the fact that the "effective" amount of time to complete a single discrete step is C/k. Please read over my post carefully. If you wait until t = 4C/k, the displacement will be 98% of the displacement you get at infinite time. It won't matter significantly if you start the next discrete step then, or wait an infinite amount of time. So the nominal time for a sequence of n discrete steps will be nC/k (or 4nC/k if you prefer).

Does the number of discrete steps have any bearing on the time it takes to reach 98% displacement? Here it appears that regardless of step size the time to reach 98% displacement is constant at 4C/k, which I find a big strange since intuitively I thought that decreasing the step size would reduce the time to reach equilibrium per step.

Theoretically, if one desires to reach complete equilibrium before the proceeding to the next discrete step, then we must wait until t -> ∞, such that in essence a "perfect" quasistatic process using discrete steps must take t = ∞²?

Thank you

 

[Chestermiller]

Does the number of discrete steps have any bearing on the time it takes to reach 98% displacement? Here it appears that regardless of step size the time to reach 98% displacement is constant at 4C/k, which I find a big strange since intuitively I thought that decreasing the step size would reduce the time to reach equilibrium per step.

Actually, no. Taking a smaller step reduces the driving force for the change, such that, irrespective of the size of the step, each discrete step takes exactly the same amount of time to reach 98% displacement. You can validate this by considering your equation for x(t). See what you get when you substitute t = 4C/k into the equation.

Theoretically, if one desires to reach complete equilibrium before the proceeding to the next discrete step, then we must wait until t -> ∞, such that in essence a "perfect" quasistatic process using discrete steps must take t = ∞²?

I submit that if you wait only 4C/k in each discrete step before starting the next step (rather than waiting an infinite amount of time), you will get results for the total work and the work to overcome damper forces which are not significantly different from the results for waiting an infinite amount of time. If you don't believe me, try solving this case and see what you get. The point I'm making is that the "effective time" to complete each step is only on the order of C/k.

I'm running out of ideas on how to explain this any better.

Chet

 

[Red_CCF]

Actually, no. Taking a smaller step reduces the driving force for the change, such that, irrespective of the size of the step, each discrete step takes exactly the same amount of time to reach 98% displacement. You can validate this by considering your equation for x(t). See what you get when you substitute t = 4C/k into the equation.

Is this because x(t) and hence dx/dt is linearly proportional to the step size P_Ef - P_Ei, so the process is faster with larger step sizes?

I submit that if you wait only 4C/k in each discrete step before starting the next step (rather than waiting an infinite amount of time), you will get results for the total work and the work to overcome damper forces which are not significantly different from the results for waiting an infinite amount of time. If you don't believe me, try solving this case and see what you get. The point I'm making is that the "effective time" to complete each step is only on the order of C/k.

I'm running out of ideas on how to explain this any better.

Chet

What is the actual penalty of not doing the process to full completion (t = ∞)? The only one I can think of is that the final desired pressure at each step is not reached, so performing to 4C/k only get to 98% of P_ef-P_ei in each step. Does this affect the reversibility of the process? For instance if we perform the process with n = ∞ steps, does performing the process in t = 4nC/k create irreversibilities that wouldn't exist otherwise?

With regards to quasistatic processes in general, the model we had assumed an insulated system. Is it correct to say that an insulated system with internal irreversibilities cannot perform a thermodynamic cycle due to the fact that system entropy can never return to its original value?

Thank you

 

[Chestermiller]

Is this because x(t) and hence dx/dt is linearly proportional to the step size P_Ef - P_Ei, so the process is faster with larger step sizes?

Yes. That's the driving force.

What is the actual penalty of not doing the process to full completion (t = ∞)?

In the limit of n→∞, t = 4nC/k is also infinite. Anyhow, I meant to say that, in the very last discrete step, you let the system fully equilibrate.

The only one I can think of is that the final desired pressure at each step is not reached, so performing to 4C/k only get to 98% of P_ef-P_ei in each step. Does this affect the reversibility of the process?

Not in the limit of n→∞; then, you achieve the same value for the work (i.e., the reversible work).

For instance if we perform the process with n = ∞ steps, does performing the process in t = 4nC/k create irreversibilities that wouldn't exist otherwise?

Not in the limit of n→∞.

With regards to quasistatic processes in general, the model we had assumed an insulated system. Is it correct to say that an insulated system with internal irreversibilities cannot perform a thermodynamic cycle due to the fact that system entropy can never return to its original value?

If the system is insulated, even if the process is carried out reversibly, you can't achieve a cycle. The most you can do is backtrack reversibly to the original state, with no net work done. In that case, you will regain the original value of the entropy. But, if there are irreversibilities, you won't even be able to get back to the initial state.

Chet

 

[Red_CCF]

In the limit of n→∞, t = 4nC/k is also infinite. Anyhow, I meant to say that, in the very last discrete step, you let the system fully equilibrate.

Not in the limit of n→∞; then, you achieve the same value for the work (i.e., the reversible work).

Not in the limit of n→∞.

So if we perform n = finite cycles at t = 4C/k, we would not get the desired final pressure until the last step (which is on purposely made bigger?) but in the limit of n→∞, since we are performing ∞ steps at t = 4C/k the "loss" at each step is made up for?

With regards to C and k, in a real spring-damper system these are physical properties, but do their values depend on how the process is performed when using them to model a piston compression? I ask because I was wondering about the 4mk/C² = small condition (assuming piston has mass) to prevent piston oscillation in each step.

If the system is insulated, even if the process is carried out reversibly, you can't achieve a cycle. The most you can do is backtrack reversibly to the original state, with no net work done. In that case, you will regain the original value of the entropy. But, if there are irreversibilities, you won't even be able to get back to the initial state.

Chet

How come reversible backtracking is not considered a cycle? I had thought that a cycle is just a series of processes that brings the system back to the initial state?

Thank you

 

[Chestermiller]

So if we perform n = finite cycles at t = 4C/k, we would not get the desired final pressure until the last step (which is on purposely made bigger?) but in the limit of n→∞, since we are performing ∞ steps at t = 4C/k the "loss" at each step is made up for?

You don't need to ask me to answer this question. This is something you can do for yourself. Redo the analysis for n=2, starting the second step at t = 4C/k rather than at ∞ and compare it with what you got when you let the first step go all the way to equilibrium. Doing this reanalysis should only take you about 5 minutes.

With regards to C and k, in a real spring-damper system these are physical properties, but do their values depend on how the process is performed when using them to model a piston compression? I ask because I was wondering about the 4mk/C² = small condition (assuming piston has mass) to prevent piston oscillation in each step.

Before we throw the piston mass back into muddy the water, let's get the cases of zero piston mass straightened out.

How come reversible backtracking is not considered a cycle? I had thought that a cycle is just a series of processes that brings the system back to the initial state?

I guess I would consider adiabatic reversible backtracking a degenerate cycle, since no work would be done and no heat would be transferred.

 

[Red_CCF]

You don't need to ask me to answer this question. This is something you can do for yourself. Redo the analysis for n=2, starting the second step at t = 4C/k rather than at ∞ and compare it with what you got when you let the first step go all the way to equilibrium. Doing this reanalysis should only take you about 5 minutes.

From previous results for n = 2 where both stages go from t = 0 to infinity

$W_{spring} = \frac{P_{Ef}^2-P_{Ei}^2}{2k}$
$W_{damper} = \frac{(P_{Ef}-P_{Ei})^2}{4k}$

For the new first step from t = 0 to 4C/k and second step for t = 0 to infinity (assuming we go to equilibrium in the second step and not to 4C/k)

First step:
$W_{spring} = 0.982\frac{(P_{Ef}+P_{Ei})^2-4P_{Ei}^2}{8k}$
$W_{damper} = 0.982\frac{(P_{Ef}-P_{Ei})^2}{8k}$

I'm assuming the second step starts at 0.982(P_{Ef}+P_{Ei})/2 where the first step left off to P_{Ef}

Second step:
$W_{spring} = \frac{4P_{Ef}^2-0.982^2(P_{Ef}+P_{Ei})^2}{8k}$
$W_{damper} = \frac{(0.509P_{Ef}-0.491P_{Ei})^2}{2k}$

Sum

$W_{spring} = \frac{P_{Ef}^2-0.982P_{Ei}^2-0.00443(P_{Ef}+P_{Ei})}{2k}$
$W_{damper} = \frac{1.01P_{Ef}^2+0.974P_{Ei}^2-1.9816P_{Ef}P_{Ei}}{4k}$

So it looks like the difference is:

$W_{diff,spring} = \frac{-0.0177P_{Ei}^2+0.00443(P_{Ef}+P_{Ei})}{2k}$
$W_{diff,damper} = \frac{0.01P_{Ef}^2+0.026P_{Ei}^2-0.0184P_{Ef}P_{Ei}}{4k}$

Numerically they do not appear to be too significant, since the coefficients are all < 0.03.

Thank you

 

[Chestermiller]

From previous results for n = 2 where both stages go from t = 0 to infinity

$W_{spring} = \frac{P_{Ef}^2-P_{Ei}^2}{2k}$
$W_{damper} = \frac{(P_{Ef}-P_{Ei})^2}{4k}$

As you pointed out in a recent post, we have been missing a factor of A² in these equations for the work.

For the new first step from t = 0 to 4C/k and second step for t = 0 to infinity (assuming we go to equilibrium in the second step and not to 4C/k)

First step:
$W_{spring} = 0.982\frac{(P_{Ef}+P_{Ei})^2-4P_{Ei}^2}{8k}$
$W_{damper} = 0.982\frac{(P_{Ef}-P_{Ei})^2}{8k}$

I'm assuming the second step starts at 0.982(P_{Ef}+P_{Ei})/2 where the first step left off to P_{Ef}

I'm proud of you for attacking this problem so aggressively, but this assumption is not correct. The second step starts out at (P_{Ef}+P_{Ei})/2 where the first step left off; it is the displacement x that starts the second step at a smaller value by 1.8%. In the second step, the displacement will be 1.8% larger than before.

Second step:
$W_{spring} = \frac{4P_{Ef}^2-0.982^2(P_{Ef}+P_{Ei})^2}{8k}$
$W_{damper} = \frac{(0.509P_{Ef}-0.491P_{Ei})^2}{2k}$

Sum

$W_{spring} = \frac{P_{Ef}^2-0.982P_{Ei}^2-0.00443(P_{Ef}+P_{Ei})}{2k}$
$W_{damper} = \frac{1.01P_{Ef}^2+0.974P_{Ei}^2-1.9816P_{Ef}P_{Ei}}{4k}$

So it looks like the difference is:

$W_{diff,spring} = \frac{-0.0177P_{Ei}^2+0.00443(P_{Ef}+P_{Ei})}{2k}$
$W_{diff,damper} = \frac{0.01P_{Ef}^2+0.026P_{Ei}^2-0.0184P_{Ef}P_{Ei}}{4k}$

Numerically they do not appear to be too significant, since the coefficients are all < 0.03.

It is easier to do this analysis by determining the change in the total amount of work, rather than the individual contributions of spring and damper. During the first step, the pressure is constant at (P_{Ef}+P_{Ei})/2 , and during the second step, the pressure is constant at P_{Ei}. But the displacement is smaller in the first step, and larger in the second step. The total final displacement will be the same as before. The change in the total amount of work will be all added damper dissipation work. See what you can come up with by trying this simpler approach. It should be proportional ato (P_{Ef}-P_{Ei})^2.

Chet

 

[Chestermiller]

In my previous reply, I meant to say that the pressure in the secomd step is PEf, rather than PEi.

Chet

 

[Red_CCF]

I'm proud of you for attacking this problem so aggressively, but this assumption is not correct. The second step starts out at (P_{Ef}+P_{Ei})/2 where the first step left off; it is the displacement x that starts the second step at a smaller value by 1.8%. In the second step, the displacement will be 1.8% larger than before.

I got the pressure terms mixed up and forgot that the pressure terms are applied pressure and not related to the cylinder gas pressure.

It is easier to do this analysis by determining the change in the total amount of work, rather than the individual contributions of spring and damper. During the first step, the pressure is constant at (P_{Ef}+P_{Ei})/2 , and during the second step, the pressure is constant at P_{Ei}. But the displacement is smaller in the first step, and larger in the second step. The total final displacement will be the same as before. The change in the total amount of work will be all added damper dissipation work. See what you can come up with by trying this simpler approach. It should be proportional ato (P_{Ef}-P_{Ei})^2.

Chet

For total work I got:

$W = \frac{A^2}{4k}[(e^{-4}-3)P_{Ef}^2-(e^{-4}-1)P_{Ei}^2+2P_{Ei}P_{Ef}]$
and the extra damper work should be
$e^{-4}P_{Ef}^2-e^{-4}P_{Ei}^2$

What is this extra work used to overcome?

Thank you

 

[Chestermiller]

For total work I got:

$W = \frac{A^2}{4k}[(e^{-4}-3)P_{Ef}^2-(e^{-4}-1)P_{Ei}^2+2P_{Ei}P_{Ef}]$
and the extra damper work should be
$e^{-4}P_{Ef}^2-e^{-4}P_{Ei}^2$

This is not what I got. I worked it this way:

W =A (P₁Δx₁ + P₂Δx₂)

where
$$P_1=\frac{(P_{Ef}+P_{Ei})}{2}$$
$$P_2=P_{Ef}$$
$$Δx_1=A\frac{(P_{Ef}-P_{Ei})}{2k}(1-e^{-4})$$
$$Δx_2=A\frac{(P_{Ef}-P_{Ei})}{2k}(1+e^{-4})$$

For the extra damper work, I got:

$$A^2\frac{(P_{Ef}-P_{Ei})^2}{4k}e^{-4}$$

What is this extra work used to overcome?

This extra work does not overcome anything. It is additional work that has to be done to deform the damper, which dissipates more mechanical energy because the velocity of the piston is higher throughout the second step (and the rate of dissipation of mechanical energy by the damper is proportional to its velocity squared).

Chet

 

[Red_CCF]

This is not what I got. I worked it this way:

W =A (P₁Δx₁ + P₂Δx₂)

where
$$P_1=\frac{(P_{Ef}+P_{Ei})}{2}$$
$$P_2=P_{Ef}$$
$$Δx_1=A\frac{(P_{Ef}-P_{Ei})}{2k}(1-e^{-4})$$
$$Δx_2=A\frac{(P_{Ef}-P_{Ei})}{2k}(1+e^{-4})$$

For the extra damper work, I got:

$$A^2\frac{(P_{Ef}-P_{Ei})^2}{4k}e^{-4}$$

Was the total work I calculated correct? I actually had a typo and the extra damper work should have been
$$\frac{A^2e^{-4}(P_{Ef}^2-P_{Ei}^2)}{4k}$$
which is still different. The original total work when each step takes to t = infinity was:
$$\frac{3P_{Ef}^2-P_{Ei}^2-2P_{Ef}P_{Ei}}{4k}$$
and I just found the difference between the two total work terms.

I computed this integral for total work. The first integral I did from 0 to 4C/k and the second from 0 to infinity.
$$W = ∫A(P_{Ef}+P_{Ei})/2dx + ∫AP_{Ef}dx$$
$$dx = -\frac{A(P_{Ef} - P_{Ei})}{2C}e^{-kt/C}dt$$

The e^-4 term only shows up in the first integral for me from next to the product of the exerted pressure and pressure difference from dx/dt: (P_Ef - P_Ei)(P_Ef + P_Ei) = (P_Ef² - P_Ei²) which is what I ended up with.

Another thing I found is that the total work in the second step was the same in this case as when the first step goes to completion, but it was not the case in your calculation.

This extra work does not overcome anything. It is additional work that has to be done to deform the damper, which dissipates more mechanical energy because the velocity of the piston is higher throughout the second step (and the rate of dissipation of mechanical energy by the damper is proportional to its velocity squared).

Chet

The dx/dt equation in the second step doesn't appear to be different than that where the first step is performed to completion. How come the velocity of the piston is higher?

Thank you

 

[Chestermiller]

You need to be very careful how you handle the differential equation for the second step. Don't forget that displacement from the first step is still continuing to occur when you raise the pressure in the second step. One way of getting the time-dependent displacement variation for the second step is to just treat it as a continuation of the first step. But the initial condition is not zero. Also, the differential equation during the time that the second step is occurring is:

$$\frac{dx}{dt}+\frac{k}{C}x=A\frac{(P_{Ef}-P_{Ei})}{C}$$

Note that there is no factor of 2 in the denominator on the right hand side because we are continuing the solution from the first step, and, during the second step of the deformation, the driving force has jumped up to $(P_{Ef}-P_{Ei})$. The initial condition for the second step is:
$$x\left(t=\frac{4k}{C}\right)=A\frac{(P_{Ef}-P_{Ei})}{2k}(1-e^{-4})$$

When you get the solution to this set of equations for x, you subtract the solution for x(4k/C) to get Δx₂(t).

I hope this mathematics makes sense to you. If not, I'll try to show you how to get the results in a different way.

Chet

 

[Chestermiller]

There were a couple of errors in my previous post that I want to correct. The text and equations should have read:

The initial condition for the second step is:
$$x\left(t=\frac{4C}{k}\right)=A\frac{(P_{Ef}-P_{Ei})}{2k}(1-e^{-4})$$

When you get the solution to this set of equations for x, you subtract the solution for x(4C/k) to get Δx₂(t).

Note that the corrections involved changing 4k/C to 4C/k in two places.

Chet

 

[Red_CCF]

You need to be very careful how you handle the differential equation for the second step. Don't forget that displacement from the first step is still continuing to occur when you raise the pressure in the second step. One way of getting the time-dependent displacement variation for the second step is to just treat it as a continuation of the first step. But the initial condition is not zero. Also, the differential equation during the time that the second step is occurring is:

$$\frac{dx}{dt}+\frac{k}{C}x=A\frac{(P_{Ef}-P_{Ei})}{C}$$

Note that there is no factor of 2 in the denominator on the right hand side because we are continuing the solution from the first step, and, during the second step of the deformation, the driving force has jumped up to $(P_{Ef}-P_{Ei})$.

I'm confused about why the RHS of the ODE is not divided by 2 since the initial pressure is (P_Ei + P_Ef )/2 and the final pressure is P_Ef. Also, is x(t) and Δx(t) equivalent?

The initial condition for the second step is:
$$x\left(t=\frac{4k}{C}\right)=A\frac{(P_{Ef}-P_{Ei})}{2k}(1-e^{-4})$$

When you get the solution to this set of equations for x, you subtract the solution for x(4k/C) to get Δx₂(t).

I hope this mathematics makes sense to you. If not, I'll try to show you how to get the results in a different way.

Chet

Is initial condition equal to x₁ (4C/k)? What I am confused by is why the initial position has a (1 - e^-4) term as opposed to just e^-4 such that $x_2(0) = A\frac{(P_{Ef}-P_{Ei})}{2k}e^{-4}$. Intuitively I thought that in the limit where the first step is performed to completion, the initial position of the second step should approach 0 but in the above case it looks to approach $A\frac{(P_{Ef}-P_{Ei})}{2k}$ ?

I also noticed that the initial position of the piston for the second step is positive, is this to be consistent with the sign notation that x(t) is increasingly negative with compression and we are starting "above" zero?

After I obtain a general expression for x₂(t), I substitute the above expression as x₂(0) and integrate ∫P_Efdx from 0 to infinity to obtain the work for the second step?

Thank you

 

[Chestermiller]

I'm confused about why the RHS of the ODE is not divided by 2 since the initial pressure is (P_Ei + P_Ef )/2 and the final pressure is P_Ef. Also, is x(t) and Δx(t) equivalent?

I'm afraid I've gotten you very confused, hopefully not beyond recovery. In the previous two posts, I was trying to show how to do the integration to determine x in one fell swoop, starting from x = 0, so the second part of the deformation is just a continuation of the deformation in the first part. Even though the analysis is correct and leads to the correct result, it was not worth the confusion it has caused. So please disregard what I did in the previous two posts.

Is initial condition equal to x₁ (4C/k)? What I am confused by is why the initial position has a (1 - e^-4) term as opposed to just e^-4 such that $x_2(0) = A\frac{(P_{Ef}-P_{Ei})}{2k}e^{-4}$. Intuitively I thought that in the limit where the first step is performed to completion, the initial position of the second step should approach 0 but in the above case it looks to approach $A\frac{(P_{Ef}-P_{Ei})}{2k}$ ?

If this is the approach that works best for you (I like it better than my approach in the previous two posts), then let's continue with that. Regarding the sign, I'm using x positive if it is in the compression direction. So, the initial condition on x₂, according to your approach, would be $x_2(0) = -A\frac{(P_{Ef}-P_{Ei})}{2k}e^{-4}$, and, in the differential equation for x₂, you would include the factor of 2 in the denominator of the right hand side. So, if you solve the differential equation subject to this initial condition, what do you get?

After I obtain a general expression for x₂(t), I substitute the above expression as x₂(0) and integrate ∫P_Efdx from 0 to infinity to obtain the work for the second step?

Yes, but making sure that you make proper use of the initial condition that $x_2(0) = -A\frac{(P_{Ef}-P_{Ei})}{2k}e^{-4}$.

Sorry for all the confusion I caused.

Chet

 

[Red_CCF]

I'm afraid I've gotten you very confused, hopefully not beyond recovery. In the previous two posts, I was trying to show how to do the integration to determine x in one fell swoop, starting from x = 0, so the second part of the deformation is just a continuation of the deformation in the first part. Even though the analysis is correct and leads to the correct result, it was not worth the confusion it has caused. So please disregard what I did in the previous two posts.

If this is the approach that works best for you (I like it better than my approach in the previous two posts), then let's continue with that. Regarding the sign, I'm using x positive if it is in the compression direction. So, the initial condition on x₂, according to your approach, would be $x_2(0) = -A\frac{(P_{Ef}-P_{Ei})}{2k}e^{-4}$, and, in the differential equation for x₂, you would include the factor of 2 in the denominator of the right hand side. So, if you solve the differential equation subject to this initial condition, what do you get?


Yes, but making sure that you make proper use of the initial condition that $x_2(0) = -A\frac{(P_{Ef}-P_{Ei})}{2k}e^{-4}$.

Sorry for all the confusion I caused.

Chet

The ODE looked like it was setup for the combined process but I wasn't 100% sure. So if I am isolating for the second step as below:
$$\frac{dx}{dt}+\frac{k}{C}x=A\frac{(P_{Ef}-P_{Ei})}{2C}$$
After substituting $x_2(0) = -A\frac{(P_{Ef}-P_{Ei})}{2k}e^{-4}$ the x₂(t) I get is:
$$x_2(t) = A\frac{(P_{Ef}-P_{Ei})}{k}[1-(e^{-4}/2+1)e^{\frac{-kt}{C}})]$$
After integrating the work I get for the second stage is:
$$W_2 = \frac{A^2(P_{Ef}^2-P_{Ef}P_{Ei})}{k}(e^{-4}/2+1)$$
The work above is positive, which I'm guessing is due to the fact that compression is considered positive work? Also the x_2(t) equation is not divided by 2k as I had expected. I included the initial pressure as (PEi+PEf)/2 in the ODE and didn't find any calculation errors.

The extra damper work I got was positive and equal to the one you computed using the much simpler method:

$$A^2\frac{(P_{Ef}-P_{Ei})^2}{4k}e^{-4}$$

Thank you