Moving an adiabatic partition in an adiabatic container

In summary: Can you provide a step-by-step procedure for solving this exact problem for the case of an insulated partition?I do not understand how to solve this problem without knowing the details of the equation of state and the reversible process.
  • #36
Andrew Mason said:
Sure. It is difficult to analyse a dynamic process. You might be interested in this paper I found which is similar to an analysis of the left side of the present problem.

AM
For a reversible quasi-static process, the compressive normal stress exerted by the gas on the piston face is given by the ideal gas law: $$\sigma=\frac{RT}{v}$$where v is the molar volume of the gas. For an irreversible non-quasistatic process, Newton's law of viscosity adds the basic viscous contribution to the normal stress (see Transport Phenomena by Bird, Stewart, and Lightfoot) such that, at the piston face, $$\sigma=\frac{RT}{v}-\frac{4}{3}\frac{\mu}{v}\frac{d v}{d t}$$where ##\mu## is the gas viscosity, and where all the properties are evaluated at the piston face.

If the piston velocity is small compared to the speed of sound in the gas, the molar volume of the gas will be essentially uniform, and v in these equations can be replaced by the total gas volume V divided by the number of moles of gas n. In addition, if, as expected, the rapid irreversible deformation of the gas is turbulent, the gas viscosity in this equation can be replaced by the turbulent viscosity ##\mu_T## which is relatively independent of temperature. Under these circumstances, the above equation reduces to:
$$\sigma=\frac{nRT}{V}-\frac{4}{3}\frac{\mu_T}{V}\frac{d V}{d t}$$
So, as a result of these considerations, the compressive normal stresses exerted by the gases on the left and right faces of the partition in the present problem are given by:$$\sigma_L=\frac{n_LRT_L}{V_L}-\frac{4}{3}\frac{\mu_T}{V_L}\frac{d V_L}{d t}$$and $$\sigma_R=\frac{n_RRT_R}{V_R}-\frac{4}{3}\frac{\mu_T}{V_R}\frac{d V_R}{d t}=\frac{n_RRT_R}{V_R}+\frac{4}{3}\frac{\mu_T}{V_R}\frac{d V_L}{d t}$$where use has been made here of the fact that the sum of the two gas volumes is constant.

Since the piston is massless and frictionless, we have that $$\sigma_L=\sigma_R=\sigma$$and thus
$$\sigma=\frac{n_LRT_L}{V_L}-\frac{4}{3}\frac{\mu_T}{V_L}\frac{d V_L}{d t}=\frac{n_RRT_R}{V_R}+\frac{4}{3}\frac{\mu_T}{V_R}\frac{d V_L}{d t}$$Eliminating ##dV_L/dt##from these relationshiips yields $$\sigma=\frac{n_LRT_L+n_RRT_R}{(V_L+V_R)}\tag{1}$$ But, since the internal energy of the combined system is constant, we have $$n_LT_L+n_RT_R=n_LT_{Li}+n_RT_{Ri}$$and Eqn. 1 becomes:$$\sigma=\frac{n_LRT_{Li}+n_RRT_{R_i}}{(V_{Li}+V_{Ri})}=\frac{(P_{Li}V_{Li}+P_{Ri}V_{Ri})}{(V_{Li}+V_{Ri})}=P_f\tag{2}$$where ##P_f## is the final pressure. Eqn. 2 indicates that, at all times during the irreversible process, the normal stress on both sides of the partition is approximately equal to the final pressure.
 
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  • #37
I've also calculated the entropy changes for the two chambers and the two approximations, expressed as ##\Delta S/C_v##. I found that, even though the predicted temperature changes and volume changes are very similar for the 2 approximations, the entropy changes for the two chambers and the total entropy changes are quite different.

Andrew approximation:
Left = 0.0382
Right = 0.0
Total = 0.0382

Chet approximation:
Left = 0.0162
Right = 0.0338
Total = 0.0498

As you can see, with the viscosity-based approximation that I am proposing, twice as much entropy is generated in the right compartment as in the left, compared to your approximation which features zero entropy generation in the right compartment. And the total entropy generation is greater with my proposed approximation by about 25 %.
 
  • #38
Chestermiller said:
I've also calculated the entropy changes for the two chambers and the two approximations, expressed as ##\Delta S/C_v##. I found that, even though the predicted temperature changes and volume changes are very similar for the 2 approximations, the entropy changes for the two chambers and the total entropy changes are quite different.

Andrew approximation:
Left = 0.0382
Right = 0.0
Total = 0.0382

Chet approximation:
Left = 0.0162
Right = 0.0338
Total = 0.0498

As you can see, with the viscosity-based approximation that I am proposing, twice as much entropy is generated in the right compartment as in the left, compared to your approximation which features zero entropy generation in the right compartment. And the total entropy generation is greater with my proposed approximation by about 25 %.
There are two things I am struggling with in your analysis. 1. since it is an ideal gas, I don't understand where the viscosity comes from. 2. I don't see how there could be more entropy generated on the compressed right side. Because the right side is compressed, the right side pressure effectively determines the work that is done, which is very close to a reversible/quasi-static adiabatic compression. The left side, however, has undergone a partial free expansion so entropy has increased.

Suppose that in the exact middle of the cylinder we added a fixed massless adiabatic panel that has a small aperature and put the panel a very small distance, say 1 mm, to the left of the moveable adiabatic panel (the moveable panel can not move farther to the left). Initially the space between the panels is a vacuum. We then open the aperture and gas from the left flows into the space. When the pressure in the 1mm space is a tiny bit greater than the pressure in the right side, the right panel moves. It keeps moving until the pressures in all three compartments are equal. In that case, the work done is very close to that of a quasi-static reversible adiabatic compression of the right side. So very little entropy increase on the right side. The left side, however, in expanding has done less work than ##\int P_{internal}dV## so it has more internal energy than it would have if it had followed a reversible path from its initial to final volume. Therefore, the reversible path is not adiabatic but one in which there is positive heat flow. So the entropy of the left side must increase.

AM
 
  • #39
Andrew Mason said:
There are two things I am struggling with in your analysis. 1. since it is an ideal gas, I don't understand where the viscosity comes from.
The reason you are struggling with this is that you have never had a course in Newtonian fluid dynamics. Newtonian fluid dynamics deals with the behavior of liquids and gases (including gases in the low-density limit of ideal gas behavior) at finite rates of deformation. According to Newton's law of viscosity, expressed in proper tensorial form, the state of stress in a Newtonian fluid (even in the ideal gas region) is described in Cartesian coordinates by (see Bird, Stewart, and Lightfoot, Transport Phenomena, Chapter 1):
$$\Pi_{xx}=p-2\mu\frac{\partial v_x}{\partial x}+\frac{2}{3}\mu\left(\frac{\partial v_x}{\partial x}+\frac{\partial v_y}{\partial y}+\frac{\partial v_z}{\partial z}\right)$$
$$\Pi_{yy}=p-2\mu\frac{\partial v_y}{\partial y}+\frac{2}{3}\mu\left(\frac{\partial v_x}{\partial x}+\frac{\partial v_y}{\partial y}+\frac{\partial v_z}{\partial z}\right)$$
$$\Pi_{zz}=p-2\mu\frac{\partial v_z}{\partial z}+\frac{2}{3}\mu\left(\frac{\partial v_x}{\partial x}+\frac{\partial v_y}{\partial y}+\frac{\partial v_z}{\partial z}\right)$$
$$\Pi_{xy}=-\mu\left(\frac{\partial v_x}{\partial y}+\frac{\partial v_y}{\partial x}\right)$$
$$\Pi_{xz}=-\mu\left(\frac{\partial v_x}{\partial z}+\frac{\partial v_z}{\partial x}\right)$$
$$\Pi_{yz}=-\mu\left(\frac{\partial v_y}{\partial z}+\frac{\partial v_z}{\partial y}\right)$$
where the ##\Pi's## are the components of the compressive stress tensor in a deforming liquid or gas, p is the local pressure calculated from the equation of state in terms of the local temperature and specific volume (e.g., the ideal gas law), the v's are the components of the local velocity vector (field), and ##\mu## is the fluid viscosity.
Please note that, with all due respect to what you learned about ideal gases, without viscosity, there is no mechanism for non-quasistatic mechanical entropy generation in gases. So to understand what is happening in a non-quasistatic deformation of a gas, viscosity is an essential consideration. For more details on this, please see Example 11D.1. Equation of change for entropy in Chapter 11, Transport Phenomena.

2. I don't see how there could be more entropy generated on the compressed right side. Because the right side is compressed, the right side pressure effectively determines the work that is done, which is very close to a reversible/quasi-static adiabatic compression.
This is just your own personal opinion. If what you are saying is correct, then, for a massless frictionless piston, since the normal stress exerted by the two gases on the opposite faces of the partition must always be equal, the normal stress exerted by the gas to the left on the partition starts out at ##2P_B##, but then drops immediately to ##1P_B## and then rises during the process to ##1.5P_B## (so that it matches what the stress is doing on the right of the partition) while, at the same time, the gas on the left is expanding. In my judgment, it defies logic to say that the force on the left increases while the gas expands.

The velocity of the partition is the same for both gases, and the magnitude of the rate of change of volume is also the same for both. From fluid dynamics, we know that the rate of volumetric strain is equal to the rate of volume change divided by the volume. Initially the magnitudes of the rates of volumetric strain are the same for both gases because the initial volumes are equal. This means that their rates of viscous dissipation are equal. However, as time progresses, the left volume becomes larger and the right volume becomes smaller. This means that the magnitude of the rate of volumetric strain on the right becomes larger than than the rate of volumetric strain on the left. This means that the rate of viscous dissipation on the right becomes larger than the rate of viscous dissipation on the left. So we expect a larger entropy increase on the right than on the left.
 
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  • #40
Chestermiller said:
This is just your own personal opinion. If what you are saying is correct, then, for a massless frictionless piston, since the normal stress exerted by the two gases on the opposite faces of the partition must always be equal, the normal stress exerted by the gas to the left on the partition starts out at 2PB, but then drops immediately to 1PB and then rises during the process to 1.5PB
(so that it matches what the stress is doing on the right of the partition) while, at the same time, the gas on the left is expanding. In my judgment, it defies logic to say that the force on the left increases while the gas expands.

Well, the force on the gas on the right has to increase as the gas on the left expands. So what is there to provide that force other than the left side? It is the resistance to the applied pressure that results in a force. As that resisitance increases (i.e. pressure on the right) the left side applies more force.

AM
 
  • #41
Andrew Mason said:
Well, the force on the gas on the right has to increase as the gas on the left expands. So what is there to provide that force other than the left side? It is the resistance to the applied pressure that results in a force. As that resisitance increases (i.e. pressure on the right) the left side applies more force.

AM
Ask yourself this: how can the gas on the left possibly apply more force if the gas on the left is expanding? On the other hand, consider this:

The force on the right remains relatively constant as a result of (a) the equation of state compressive stress contribution ##p_R## increasing in conjunction with (b) the viscous compressive stress decreasing toward zero (as a result of the partition slowing down). The force on the left side also remains relatively constant as a result of (a) the equation of state compressive stress contribution ##p_L## decreasing in conjunction with (b) the viscous compressive stress (which, on this side of the partition, is negative, i.e. tensile) increasing toward zero (as a result of the partition slowing down).
 
  • #42
Chestermiller said:
Ask yourself this: how can the gas on the left possibly apply more force if the gas on the left is expanding?
The expansion results in a directed flow of gas to the right. That flow carries momentum. That momentum is transferred to the gas molecules on the right. The rate of change of momentum is a force. That force compresses the right side.

AM
 
  • #43
Andrew Mason said:
The expansion results in a directed flow of gas to the right. That flow carries momentum. That momentum is transferred to the gas molecules on the right. The rate of change of momentum is a force. That force compresses the right side.

AM
I'm talking exclusively about the gas on the left. Its volume is increasing and its temperature is decreasing. How can the force that it is exerting on the partition be increasing?

There is momentum transfer on both sides of the partition. These momentum transfer effects are captured by the turbulent viscosity contributions to the total compressive stresses. The viscous contributions give rise to entropy generation both in the left and the right compartments.
 
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  • #44
Chestermiller said:
I'm talking exclusively about the gas on the left. Its volume is increasing and its temperature is decreasing. How can the force that it is exerting on the partition be increasing?
The internal energy of the gas is converted into kinetic energy of the directed flow to the right. As the pressure differential between left and right is largest initially, kinetic energy of the gas flow and temperature is greatest initially and pressure, therefore, force, is least (Bernoulli principle) in the flowing part of the gas (which is in the volume of gas nearest the partition). As the gas flow slows, kinetic energy is reduced and pressure, therefore, force increases.
There is momentum transfer on both sides of the partition. These momentum transfer effects are captured by the turbulent viscosity contributions to the total compressive stresses. The viscous contributions give rise to entropy generation in both the left and the right compartments.
There will be some oscillation but these oscillations will die down due not to friction but to randomization of the molecular motion. The paper that I mentioned earlier discusses this.

AM
 
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  • #45
Andrew Mason said:
The internal energy of the gas is converted into kinetic energy of the directed flow to the right. As the pressure differential between left and right is largest initially, kinetic energy of the gas flow and temperature is greatest initially and pressure, therefore, force, is least (Bernoulli principle) in the flowing part of the gas (which is in the volume of gas nearest the partition). As the gas flow slows, kinetic energy is reduced and pressure, therefore, force increases.
There will be some oscillation but these oscillations will die down due not to friction but to randomization of the molecular motion. The paper that I mentioned earlier discusses this.

AM
In your paper, the "dynamic pressure" is given by the equation $$\hat{P}=\frac{nRT}{xA}\left[1-v\sqrt{\frac{8M}{\pi RT}}\right]=\frac{nRT}{xA}\left[1-\frac{dx}{dt}\sqrt{\frac{8M}{\pi RT}}\right]\tag{3}$$This equation can be re-expresssed as $$\hat{P}=\frac{nRT}{V}-\frac{1}{V}\frac{dV}{dt}\frac{n}{A}\sqrt{\frac{8MRT}{\pi}}$$The quantity ##\frac{n}{A}\sqrt{\frac{8MRT}{\pi}}## in this equation is proportional to the gas viscosity approximation determined by Maxwell in 1860 using kinetic theory. Note the similarity of this expression for the "dynamic pressure" to the expression I have presented for compressive stress ##\Pi##. This is how the continuum viscosity theory connects to your kinetic theory approach. In my development, all I have done is replace your gas viscosity with turbulent viscosity to take into account the fact that the flow will be turbulent.

The net result of the approach indicated in your paper actually supports my contention that there will be significant entropy generation in both compartments.
 
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  • #46
@Andrew Mason

So, if I understand correctly, you accept the equation for the "dynamic pressure" given in the paper you have referenced, right? Because, if that is the case, I will solve the equations for our partitioned gas system (numerically), subject to this dynamic pressure representation, to obtain the gas temperatures, volumes, and dynamic pressure at the partition as a function of time. We can then see once and for all whose approximation is a better representation for predicting the final state of the system and the entropy changes. Are you willing to accept these results if I go to the trouble to obtain them?

Chet
 
  • #47
Chestermiller said:
@Andrew Mason

So, if I understand correctly, you accept the equation for the "dynamic pressure" given in the paper you have referenced, right? Because, if that is the case, I will solve the equations for our partitioned gas system (numerically), subject to this dynamic pressure representation, to obtain the gas temperatures, volumes, and dynamic pressure at the partition as a function of time. We can then see once and for all whose approximation is a better representation for predicting the final state of the system and the entropy changes. Are you willing to accept these results if I go to the trouble to obtain them?

Chet
Let me think about it. I wouldn't want to ruin the holidays for you... Merry Christmas!
AM
 
  • #48
This is an analysis of the present problem using the model of non-quasistatic gas behavior recommended by @Andrew Mason in posts #35 and #44 of the present thread. The starting equations are as follows:

Energy Balances on the two compartments:
$$nC_v\frac{dT_L}{dt}=-\hat{P}\frac{dV_L}{dt}\tag{1}$$
$$nC_v\frac{dT_R}{dt}=-\hat{P}\frac{dV_R}{dt}=+\hat{P}\frac{dV_L}{dt}\tag{2}$$where n is the number of moles of gas in each of the two compartments and ##\hat{P}## is the time-dependent "dynamic pressure" on each of the two opposite faces of the massless frictionless partition, given by the the non-quasistatic model of gas behavior in Andrew's reference:

$$\hat{P}=\frac{nRT_L}{V_L}-\frac{\mu(T_L)}{V_L}\frac{dV_L}{dt}=\frac{nRT_R}{V_R}-\frac{\mu(T_R)}{V_R}\frac{dV_R}{dt}=\frac{nRT_R}{V_R}+\frac{\mu(T_R)}{V_R}\frac{dV_L}{dt}\tag{3}$$where ##\mu(T)## is a temperature-dependent parameter with units of viscosity that is proportional to ##\sqrt{T}##.

Initial conditions (t = 0) on the problem are: $$V_L=V_R=V$$
$$T_L=2T_B$$
$$T_R=T_B$$$$P_L=2P_B=2\frac{nRT_B}{V}$$
$$P_R=P_B=\frac{nRT_B}{V}$$
 
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  • #49
This is a continuation of the analysis begun in the previous post.

We can solve Eqns. 3 (representing the non-quasistatic Mungan gas model recommended by @Andrew Mason) to obtain the time derivative of the left compartment volume ##dV_L/dt## and the "dynamic pressure" ##\hat{P}## solely in terms of the compartment temperatures, volumes, and gas viscosities:
$$\frac{dV_L}{dt}=nR\frac{(T_LV_R-T_RV_L)}
{(\mu_LV_R+\mu_RV_L)}\tag{4}$$and$$\hat{P}=nR\frac{(T_L\mu_R+T_R\mu_L)}{(\mu_LV_R+\mu_RV_L)}\tag{5}$$where ##\mu_L=\mu(T_L)## and ##\mu_R=\mu(T_R)##.

Next, if we add Eqns. 1 and 2, we obtain:$$nC_v\frac{d(T_L+T_R)}{dt}=0$$The solution to this equation, subject to the prescribed initial conditions is $$\frac{T_L+T_R}{2}=\bar{T}=\frac{3}{2}T_B\tag{6}$$where ##\bar{T}## is the (constant) average of the two compartment temperatures throughout the deformation.
Next, if we subtract Eqn. 2 from Eqn. 1, we obtain:$$nC_v\frac{d(T_L-T_R)}{dt}=2\hat{P}\frac{dV}{dt}$$or, equivalently, $$nC_v\frac{d(T_L-T_R)}{dV_L}=-2\hat{P}\tag{7}$$At this juncture, we define the following substitutions: $$V_L=V(1+\xi_V)\tag{8a}$$
$$V_R=V(1-\xi_V)\tag{8b}$$
$$T_L=\bar{T}(1+\xi_T)\tag{8c}$$
$$T_R=\bar{T}(1-\xi_T)\tag{8d}$$
$$\mu_L=\mu(\bar{T})\sqrt{1+\xi_T}\tag{8e}$$
$$\mu_R=\mu(\bar{T})\sqrt{1-\xi_T}\tag{8e}$$where, in writing Eqns. 8e and 8f, we have made use of the fact that, in the Mungan model, ##\mu## is proportional to ##\sqrt{T}##. If we substitute Eqns. 8 into Eqns. 4, 5, and 6, we obtain:$$\frac{d\xi_V}{d\tau}=\frac{2(\xi_T-\xi_V)}{\sqrt{1+\xi_T}(1-\xi_V)+\sqrt{1-\xi_T}(1+\xi_V)}\tag{9}$$
$$\hat{P}=\frac{nR\bar{T}}{V}\left[\frac{\sqrt{1+\xi_T}(1-\xi_T)+\sqrt{1-\xi_T}(1+\xi_T)}{\sqrt{1+\xi_T}(1-\xi_V)+\sqrt{1-\xi_T}(1+\xi_V)}\right]\tag{10}$$
$$\frac{d\xi_T}{d\xi_V}=-(\gamma - 1)\left[\frac{\sqrt{1+\xi_T}(1-\xi_T)+\sqrt{1-\xi_T}(1+\xi_T)}{\sqrt{1+\xi_T}(1-\xi_V)+\sqrt{1-\xi_T}(1+\xi_V)}\right]\tag{11}$$
where the dimensionless time ##\tau## is given by ##\tau=\frac{nR\bar{T}}{V\mu(\bar{T})}t##. For the present problem, initial conditions of these ordinary differential equations are: $$\xi_V(0)=0$$
and $$\xi_T(0)=\frac{1}{3}$$
One will note from Eqn. 11 that, even though this is a time-dependent problem, the trajectory of the dimensionless temperature variation ##\xi_T## as a function of the dimensionless volume variation ##\xi_V## is unique, and independent of the time history. Therefore, Eqn. 11 can be solved once-and-for-all for ##\xi_T## (and the dynamic pressure ##\hat{P}##) vs ##\xi_V##. In the present analysis this has been done by integrating Eqn. 11 numerically from ##\xi_V=0## to the final equilibrium point where, according to Eqn. 9, ##\xi_T=\xi_V##. The results of these numerical calculations using the recommended Mungan model will be presented in the next post, together with corresponding predictions using Chet Miller's approximation and using Andrew Mason's approximation (to determine which approximation provides a better match to the Mungan model).
 
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  • #50
Andrew Mason (AM) APPROXIMATION
The AM approximation is based on the assumption that the gas in the right compartment suffers an ideal adiabatic reversible compression during this overall irreversible process, such that essentially all the entropy generation is confined to the left compartment (in which the gas experiences expansion). This assumption seems to be incompatible with the Mungan Model of non-quasistatitic gas deformation that AM has recommended using for more accurate calculations.

Under the AM Approximation, one has that $$T_RV_R^{(\gamma - 1)}=T_BV^{(\gamma-1)}$$and$$\hat{P}V_R^\gamma=P_BV^\gamma$$or, in terms of the dimensionless parameters defined in the present development, $$\xi_T=1-\frac{2}{3(1-\xi_V)^{(\gamma-1)}}$$and $$\frac{\hat{P}}{P_B}=\frac{1}{(1-\xi_V)^\gamma}$$The predictions from these equations will be compared with the results of the Mungan Model calculation.

Chet Miller (CM) APPROXIMATION
The CM approximation assumes that the dynamic pressure throughout the deformation is equal to the final pressure at equilibrium of the system (and also equal to the volume weighted average of the initial pressures in the two compartments). One can also show that this assumption is equivalent to neglecting the temperature-dependence of the viscosity parameter ##\mu## in the Mungan Model.

Under the CM Approximation, for the parameter values in the present problem, one has that $$\frac{\hat{P}}{P_B}=1.5$$and, from Eqns. 10 and 11 of the present development, $$\xi_T=\frac{1}{3}-(\gamma-1)\xi_V$$As with the AM model, the predictions from these equations will be compared with the results of the Mungan Model calculation.

RESULTS OF CALCULATIONS

The figure below shows the results of the Mungan Model calculations for the dimensionless temperature as a function of the dimensionless volume. Also shown in the figure, for comparison, are the predictions using the AM and CM approximations.

1577891966295.png


The results from the CM approximation are barely distinguishable for the Mungan Model results. On the other hand, there is a significant difference between the AM approximation and the Mungan Model. This suggests that the AM approximation is a poor representation of physical reality.

The figure below shows the results of the Mungan Model calculations for the dimensionless "dynamic pressure" as a function of the dimensionless volume. Also shown in the figure, for comparison, are the predictions using the AM and CM approximations.
1577896536368.png

Here again, the CM approximation provides a fairly good match to the AM-recommended Mungan Model results. On the other hand, the AM approximation deviates significantly from the Mungan Model.

The obvious conclusion here is that it is a poor approximation to assume that the gas in the low-pressure (right) compartment suffers a nearly reversible adiabatic compression in this system and that there is negligible entropy generation in the right compartment. According to the Mungan Model and the CM approximation, there is substantial irreversibility and entropy generation in both compartments.
 
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