Question about Flow between Parallel Plates

[Red_CCF]

Yes.

My original confusion arises from this. I have seen graphs in books such at the one attached, where the P-v curve for an irreversible expansion lies below that of the reversible (in the attached graph the dotted line is the reversible process). Is there something I'm missing, because the derivations above challenges this.

Also, if wall friction was neglected but viscous dissipation is now the only irreversibility, qualitatively would the trend in the attached graph be accurate since τ_xx is negative during expansion such that σ_I decreases compared to the reversible case?

The integral of the TdS equation is valid for any process, reversible or not because the changes in S, U, and V depend only on the beginning and end equilibrium states. Since the TdS equation describes differential changes between closely neighboring equilibrium states, the integral implies moving along a reversible path.

Yes, that's exactly what I've been trying to say, although you articulated it much better.
Chet

With regards to the ΔS equation for ideal gases:

ΔS=n(C_pln(T(V)/T_0)+Rln(P_0/P(V)))

For our example, since the process is quasistatic, a path of equilibrium states exists. Would integrating dS = c_p dT/T - R dp/p directly as opposed to using the above equation give the same result? Does the assumption of δQ = TdS in the equation (as opposed to δQ ≤ TdS in reality) impact this in anyway?

Is P(V) here supposed to be the interface pressure? In my substitution I used the gas pressure, which is < P_ext; should I have added F/A to P(V)?

Is the equation in this form valid for combustion in an isolated system, or must \Sigmaμ_iN_i be appended to the end of this equation?

Thank you very much

 

[rkmurtyp]

I was referred to this thread by Chet.

This is too general a question. After you've worked a few problems, you will recognize it right away.

OK. Here we go.

Here are the assumptions I'm proposing to use:

1. During the expansion or compression, the friction force on the piston is constant at F.
2. The piston is massless, and has zero heat capacity
3. All the heat generated between the wall and the piston ends up in the gas
4. The expansion and compression are carried out quasi-statically

Are you in agreement with these assumptions?
Are there any others that you think we should be using?

Which do you want to do first, expansion or compression?
What to you want included in the system: (a) just the gas or (b) the gas plus the piston? (I'm suggesting we eventually do it both ways to see how the two analyses compare).

Chet

The points 1 and 4 above need a comment.

1. Since we are interested in a process of irreversible adiabatic expansion (or compression) of an ideal gas, it is not necessary that the external pressure (Force on the pistion) be constsnt. The necessary and sufficient condition is that 1. for expansion the external pressure, P_ext plus F/A be less than the pressure of the gas, P_sym and 2. for compression the external pressure, P_extis greater than the pressure of the gas, P_symplus F/A. F is the frictional force and A the area of the pistion over which F acts.

4. Irreversible adiabatic expansion (or compression) involving friction and quasistatic process don't go together - they are mutually exclusive. If we assume presence of friction, then quasistatic process is impossible; similarly, if we assume quasistatic process absence of friction is implied.

With these in mind we discuss below, the calculation of entropy change for an irreversible adiabatic expansion (or compression) process AB of an ideal gas.

Irreversible adiabatic expansion process AB of an ideal gas.

Let the system (a fixed amount of ideal gas) go from an initial equilibrium state A (P_A, V_A, T_A) to a final equilibrium state B (P_B, V_B, T_B) by an irreversible adiabatic process involving friction. Note that presence of friction is not a necessary condition for irreversible expansion of an ideal gas. The entropy of the gas changes from S_A to S_B, The change in the entropy of the system, ΔS_sym =S_B-S_A. During this change the surroundings suffer no entropy change. ΔS_sur=0. ∴ ΔS_sym = ΔS_AB = ΔS_univ>0.

If the gas had expanded revrsibly (without friction) from state A through the same volume change, it would have reached equilibrium state C (P_C, V_C, T_C), where V_B=V_C and T_B>T_C.

From state B let us cool the system reversibly at constant volume to state C. This process requires a regenerator - an ideal device of a set of heat reservoirs (HRs) of continuously varying temperature between any two desired values of temperatures. Since this process BC is reversible ΔS_univ=0. ∴ ΔS_BC=-ΔS_CB= - ΔS_sur= -∑(Q_i/T_i), where Q_i is the heat change suffered (absorbed) by HR at temperature T_i.

Since states A and C lie on a reversible adiabat, entropy values of the system in these two states is the same ie S_A=S_C. ∴ ΔS_BA=-ΔS_AB =ΔS_BC= -∑(Q_i/T_i) <0. ∴ ΔS_AB ∑(Q_i/T_i > 0.

An alternate way of seeing this is: When the system reached state B from state A by a spontaneous (irreversible) process, we take the system back from state B to state A by a reversible path (process) BCA, BC being a reversible cooling process and CA being a reversible adiabatic compression process, thus completing a cycle. For such a cyclic process ΔS_sym=0. ∴ ΔS_AB=-ΔS_BCA=ΔS_sur= ∑(Q_i/T_i>0.

Calculation of ΔS for an irreversible adiabatic compression process AB of an ideal gas can be carried out on similar lines.

 

[Red_CCF]

For a quasistatic (reversible) process we need to satisfy the condition P_ext= P_sym. When friction is present it is impossible to satisfy this condition.

This was the first time I came across the fact that processes with friction cannot be quasistatic as everything I've read up to this point (such as the sources below and other threads) including has indicated that it is possible to have a quasistatic process with friction, so I'm having a tough time accepting this. If the argument for friction applies, what about a process with heat transfer with finite temperature gradient; can those be quasistatic then?

From Wikipedia: http://en.wikipedia.org/wiki/Quasistatic_process

An example of a quasistatic process that is not reversible is a compression against a system with a piston subject to friction — although the system is always in thermal equilibrium, the friction ensures the generation of dissipative entropy, which directly goes against the definition of reversible.

From a journal article (I only have access to the abstract unfortunately): http://adsabs.harvard.edu/abs/1960AmJPh..28..119T

Quasi-static processes are not reversible when sliding friction forces are present. An example is considered consisting of a cylinder containing a gas and equipped with a piston for which sliding friction forces are significant.

The speed (or time rate of change) of a process (slow enough of fast enough etc) does not and should not enter thermodynamic arguments - time has no role to play in (equilibrium) thermodynamics.

A process that takes the system through a series of continuous set of equilibrium states has no chance to be out of equilibrium ever, consequently be irreversible ever!

I've always thought of the slow enough argument to be qualitative to mean that the rate in which the process proceeds << the rate in which equilibrium is established during the process after an infinitessimal disturbance (an infinitessimal movement of the cylinder). Quantitatively I do not believe time ever factors into any analysis.

Irreversible adiabatic expansion process AB of an ideal gas.

Let the system (a fixed amount of ideal gas) go from an initial equilibrium state A (P_A, V_A, T_A) to a final equilibrium state B (P_B, V_B, T_B) by an irreversible adiabatic process involving friction. Note that presence of friction is not a necessary condition for irreversible expansion of an ideal gas. The entropy of the gas changes from S_A to S_B, The change in the entropy of the system, ΔS_sym =S_B-S_A. During this change the surroundings suffer no entropy change. ΔS_sur=0. ∴ ΔS_sym = ΔS_AB = ΔS_univ>0.

If the gas had expanded revrsibly (without friction) from state A through the same volume change, it would have reached equilibrium state C (P_C, V_C, T_C), where V_B=V_C and T_B>T_C.

From state B let us cool the system reversibly at constant volume to state C. This process requires a regenerator - an ideal device of a set of heat reservoirs (HRs) of continuously varying temperature between any two desired values of temperatures. Since this process BC is reversible ΔS_univ=0. ∴ ΔS_BC=-ΔS_CB= - ΔS_sur= -∑(Q_i/T_i), where Q_i is the heat change suffered (absorbed) by HR at temperature T_i.

Since states A and C lie on a reversible adiabat, entropy values of the system in these two states is the same ie S_A=S_C. ∴ ΔS_BA=-ΔS_AB =ΔS_BC= -∑(Q_i/T_i) <0. ∴ ΔS_AB ∑(Q_i/T_i > 0.

An alternate way of seeing this is: When the system reached state B from state A by a spontaneous (irreversible) process, we take the system back from state B to state A by a reversible path (process) BCA, BC being a reversible cooling process and CA being a reversible adiabatic compression process, thus completing a cycle. For such a cyclic process ΔS_sym=0. ∴ ΔS_AB=-ΔS_BCA=ΔS_sur= ∑(Q_i/T_i>0.

Calculation of ΔS for an irreversible adiabatic compression process AB of an ideal gas can be carried out on similar lines.

With regards to ΔS_BC=-ΔS_CB= - ΔS_sur= -∑(Qi/Ti); Qi here is w.r.t. the surroundings and hence positive? Other than that everything makes sense.

Thanks very much

 

[Chestermiller]

I was referred to this thread by Chet.


Thanks rkmurtyp.

The points 1 and 4 above need a comment.

1. Since we are interested in a process of irreversible adiabatic expansion (or compression) of an ideal gas, it is not necessary that the external pressure (Force on the pistion) be constsnt. The necessary and sufficient condition is that 1. for expansion the external pressure, P_ext plus F/A be less than the pressure of the gas, P_sym and 2. for compression the external pressure, P_extis greater than the pressure of the gas, P_symplus F/A. F is the frictional force and A the area of the pistion over which F acts.

In the statement of our problem, we were very specific about what we were solving. In particular, we were carrying out the expansion or compression under conditions where the piston is moved extremely slowly (quasi statically), in which case the > and < in the above comment become = signs.

4. Irreversible adiabatic expansion (or compression) involving friction and quasistatic process don't go together - they are mutually exclusive. If we assume presence of friction, then quasistatic process is impossible; similarly, if we assume quasistatic process absence of friction is implied.

We were also very specific about what we considered our system. We chose to regard the system as the combination of cylinder, piston, and gas. For this system, the expansion is adiabatic and quasistatic, since we also assume that all the heat generated at the interface between the piston and the cylinder remains within the system, and goes into the gas.

If was had chosen as our system the gas alone, then the situation would have been quite different. For this system, the process would have been non-adiabatic and non-isothermal, but reversible.

With these in mind we discuss below, the calculation of entropy change for an irreversible adiabatic expansion (or compression) process AB of an ideal gas.

Irreversible adiabatic expansion process AB of an ideal gas.

Let the system (a fixed amount of ideal gas) go from an initial equilibrium state A (P_A, V_A, T_A) to a final equilibrium state B (P_B, V_B, T_B) by an irreversible adiabatic process involving friction. Note that presence of friction is not a necessary condition for irreversible expansion of an ideal gas. The entropy of the gas changes from S_A to S_B, The change in the entropy of the system, ΔS_sym =S_B-S_A. During this change the surroundings suffer no entropy change. ΔS_sur=0. ∴ ΔS_sym = ΔS_AB = ΔS_univ>0.

If the gas had expanded revrsibly (without friction) from state A through the same volume change, it would have reached equilibrium state C (P_C, V_C, T_C), where V_B=V_C and T_B>T_C.

From state B let us cool the system reversibly at constant volume to state C. This process requires a regenerator - an ideal device of a set of heat reservoirs (HRs) of continuously varying temperature between any two desired values of temperatures. Since this process BC is reversible ΔS_univ=0. ∴ ΔS_BC=-ΔS_CB= - ΔS_sur= -∑(Q_i/T_i), where Q_i is the heat change suffered (absorbed) by HR at temperature T_i.

Since states A and C lie on a reversible adiabat, entropy values of the system in these two states is the same ie S_A=S_C. ∴ ΔS_BA=-ΔS_AB =ΔS_BC= -∑(Q_i/T_i) <0. ∴ ΔS_AB ∑(Q_i/T_i > 0.

An alternate way of seeing this is: When the system reached state B from state A by a spontaneous (irreversible) process, we take the system back from state B to state A by a reversible path (process) BCA, BC being a reversible cooling process and CA being a reversible adiabatic compression process, thus completing a cycle. For such a cyclic process ΔS_sym=0. ∴ ΔS_AB=-ΔS_BCA=ΔS_sur= ∑(Q_i/T_i>0.

Calculation of ΔS for an irreversible adiabatic compression process AB of an ideal gas can be carried out on similar lines.

In my judgement, all this is entirely equivalent to what we did in determining the change in entropy of the system. However, it is of value in providing additional information for Red_CCF of how what he calls the "TdS equations" were arrived at.

Chet

 

[Chestermiller]

My original confusion arises from this. I have seen graphs in books such at the one attached, where the P-v curve for an irreversible expansion lies below that of the reversible (in the attached graph the dotted line is the reversible process). Is there something I'm missing, because the derivations above challenges this.

I don't see an attached graph. But, as we discussed previously, there is considerable ambiguity as to what the P-v curve for an irreversible expansion lies, since, typically, in an irreversible expansion, the pressure varies with position throughout the system during an irreversible path.

Also, if wall friction was neglected but viscous dissipation is now the only irreversibility, qualitatively would the trend in the attached graph be accurate since τ_xx is negative during expansion such that σ_I decreases compared to the reversible case?

Again, I don't see the graph.

With regards to the ΔS equation for ideal gases:

ΔS=n(C_pln(T(V)/T_0)+Rln(P_0/P(V)))

For our example, since the process is quasistatic, a path of equilibrium states exists. Would integrating dS = c_p dT/T - R dp/p directly as opposed to using the above equation give the same result?

Those are exact differentials on the right hand side of the equation. So, if you integrate the dS equation between the initial and final states, you end up with the ΔS equation.

Does the assumption of δQ = TdS in the equation (as opposed to δQ ≤ TdS in reality) impact this in anyway?

Well, for the combined system of cylinder, piston, and gas, δQ is equal to zero. So certainly, that wouldn't give us the correct ΔS. See my next answer below.

Is P(V) here supposed to be the interface pressure? In my substitution I used the gas pressure, which is < P_ext; should I have added F/A to P(V)?

The above equations apply exclusively to an ideal gas between equilibrium states. Since there is no change in the piston entropy, we were able to get the change in entropy of the combined system by focusing exclusively on the gas. So for that calculation, the gas pressure would be the correct one to use. If we had used the gas as our system in the first place, then there would have been heat transferred from the piston to the gas, and the gas expansion in our process could have been considered reversible, since the heat was added in vanishingly small increments. So, if we integrated over the exact path of the gas, rather than any other arbitrary path, we would have ended up with the exact same change in entropy.

Is the equation in this form valid for combustion in an isolated system, or must \Sigmaμ_iN_i be appended to the end of this equation?

No. The equation in this form is applicable to an ideal gas without reaction. The case with reaction must take into account the changes in entropy as a result of the reaction. That's beyond what we are talking about here.

Chet

 

[rkmurtyp]

This was the first time I came across the fact that processes with friction cannot be quasistatic as everything I've read up to this point (such as the sources below and other threads) including has indicated that it is possible to have a quasistatic process with friction, so I'm having a tough time accepting this. If the argument for friction applies, what about a process with heat transfer with finite temperature gradient; can those be quasistatic then?

From Wikipedia: http://en.wikipedia.org/wiki/Quasistatic_process


From a journal article (I only have access to the abstract unfortunately): http://adsabs.harvard.edu/abs/1960AmJPh..28..119T


I've always thought of the slow enough argument to be qualitative to mean that the rate in which the process proceeds << the rate in which equilibrium is established during the process after an infinitessimal disturbance (an infinitessimal movement of the cylinder). Quantitatively I do not believe time ever factors into any analysis.


With regards to ΔS_BC=-ΔS_CB= - ΔS_sur= -∑(Qi/Ti); Qi here is w.r.t. the surroundings and hence positive? Other than that everything makes sense.

Thanks very much

The references imply this: For quasistatic processes the difference in the values of the intensive properties of the system and surrounding differ by infinitesimal amounts. To that infinitesimal extent the process would be irreversible; only in the case the differences are zero reversibility is attained - when not zero no reversibility (one may coin hundred names such as quasistatic etc, they all continue to remain irreversible). I have used in my arguments reversibility and quasistatic synonimously and ΔP to be zero ( not tending to zero!).

'Rverersibility' concept is a beatiful mental costruct that aids in simplifying the concepts and helps understanding - just as massless, frictionless pistons and pulleys help understanding the concepts easily.


I've always thought of the slow enough argument to be qualitative to mean that the rate in which the process proceeds << the rate in which equilibrium is established during the process after an infinitessimal disturbance (an infinitessimal movement of the cylinder). Quantitatively I do not believe time ever factors into any analysis.

True, time plays no role in thermodynamic processes.


With regards to ΔS_BC=-ΔS_CB= - ΔS_sur= -∑(Qi/Ti); Qi here is w.r.t. the surroundings and hence positive? Other than that everything makes sense.

You are right; Qi here is w.r.t. the surroundings and hence positive - remember when HRs receive heat Qi is positive, and negative when they lose heat.

 

[Red_CCF]

The references imply this: For quasistatic processes the difference in the values of the intensive properties of the system and surrounding differ by infinitesimal amounts. To that infinitesimal extent the process would be irreversible; only in the case the differences are zero reversibility is attained - when not zero no reversibility (one may coin hundred names such as quasistatic etc, they all continue to remain irreversible). I have used in my arguments reversibility and quasistatic synonimously and ΔP to be zero ( not tending to zero!).

'Rverersibility' concept is a beatiful mental costruct that aids in simplifying the concepts and helps understanding - just as massless, frictionless pistons and pulleys help understanding the concepts easily.

I have always thought that quasistatic processes are one where the perturbations are infinitessimal (i.e. an infinitessimally higher P during compression or a temperature gradient where the temperature differences are infinitessimal). If, say for piston-cylinder compression, ΔP = 0 and Pext and the system pressure equal exactly, I am not sure physically how the piston would move to begin with, since the net force would be exactly zero on a still piston.

Thanks very much

 

[Red_CCF]

I don't see an attached graph. But, as we discussed previously, there is considerable ambiguity as to what the P-v curve for an irreversible expansion lies, since, typically, in an irreversible expansion, the pressure varies with position throughout the system during an irreversible path.

Again, I don't see the graph.

My apologies, the graph should (hopefully) be visible now.

Those are exact differentials on the right hand side of the equation. So, if you integrate the dS equation between the initial and final states, you end up with the ΔS equation.

Well, for the combined system of cylinder, piston, and gas, δQ is equal to zero. So certainly, that wouldn't give us the correct ΔS. See my next answer below.

So to summarize, the integral of dS = c_p dT/T - Rdp/p to give the ΔS equation assumes a reversible path between initial and final states. Had I substituted p = P(V) and T = T(V) such that dS = c_p dT/T(V) - Rdp/p(V) and then integrated, I should not expect to get the same result as substituting T(V) and P(V) into the ΔS equation (as we had done earlier) as dS ≠ δQ/T?

The above equations apply exclusively to an ideal gas between equilibrium states. Since there is no change in the piston entropy, we were able to get the change in entropy of the combined system by focusing exclusively on the gas. So for that calculation, the gas pressure would be the correct one to use. If we had used the gas as our system in the first place, then there would have been heat transferred from the piston to the gas, and the gas expansion in our process could have been considered reversible, since the heat was added in vanishingly small increments. So, if we integrated over the exact path of the gas, rather than any other arbitrary path, we would have ended up with the exact same change in entropy.

Since the piston ΔS = 0 anyways, does this mean that had I chosen to add F/A to P(V) (essentially use P_ext instead), I should still get the same change in entropy using the ΔS equation, or does that equation assume the entire system is an ideal gas?

Thanks very much

 

[rkmurtyp]

I have always thought that quasistatic processes are one where the perturbations are infinitessimal (i.e. an infinitessimally higher P during compression or a temperature gradient where the temperature differences are infinitessimal). If, say for piston-cylinder compression, ΔP = 0 and Pext and the system pressure equal exactly, I am not sure physically how the piston would move to begin with, since the net force would be exactly zero on a still piston.

Thanks very much

This is the reason why reversibility is a mental construct, to imagine it in practice we invoke that quasistatic process and use all the jargon of infinitesimal differeneces etc.

Think for yourself how one can have massless frictionless pulleys, pistons etc to carry out expts in mechanics, then you will be able to appreciate the concept of reversibility in thermodynamics.

 

[rkmurtyp]

In the statement of our problem, we were very specific about what we were solving. In particular, we were carrying out the expansion or compression under conditions where the piston is moved extremely slowly (quasi statically), in which case the > and < in the above comment become = signs.

The 'extreamly slowly' doesn't qualify quasistatic condition; the inequalities define conditions for irreversible and equality sign defines conditions for reversible process


We were also very specific about what we considered our system. We chose to regard the system as the combination of cylinder, piston, and gas. For this system, the expansion is adiabatic and quasistatic, since we also assume that all the heat generated at the interface between the piston and the cylinder remains within the system, and goes into the gas.

By choosing the combination of cylinder, piston and gas as the system we sufficiently ill define the system (its properties) we end up no where in the analysis!

If was had chosen as our system the gas alone, then the situation would have been quite different. For this system, the process would have been non-adiabatic and non-isothermal, but reversible.

It is not clear to me what you are saying


In my judgement, all this is entirely equivalent to what we did in determining the change in entropy of the system. However, it is of value in providing additional information for Red_CCF of how what he calls the "TdS equations" were arrived at.

So far so good.

Chet

 

[Chestermiller]

My apologies, the graph should (hopefully) be visible now.

My original confusion arises from this. I have seen graphs in books such at the one attached, where the P-v curve for an irreversible expansion lies below that of the reversible (in the attached graph the dotted line is the reversible process). Is there something I'm missing, because the derivations above challenges this.

Not really. P in the figure is what we have been calling P_ext, which inherently assumes that the gas, cylinder, and piston constitute the system. On expansion, P_ext is less than the gas pressure would have been if the gas expansion were reversible. (See if you can verify this.) On compression, P_ext is greater than the gas pressure would have been if the gas expansion were reversible. (See if you can verify this.)

Also, if wall friction was neglected but viscous dissipation is now the only irreversibility, qualitatively would the trend in the attached graph be accurate since τ_xx is negative during expansion such that σ_I decreases compared to the reversible case?

In an irreversible expansion, P_ext is less than the gas pressure would have been if the gas expansion were reversible. Your interpretation of why is probably correct.

So to summarize, the integral of dS = c_p dT/T - Rdp/p to give the ΔS equation assumes a reversible path between initial and final states.

No, the ΔS equation applies to an ideal gas between two equilibrium states, irrespective of whether the actual process path was reversible or not.

Had I substituted p = P(V) and T = T(V) such that dS = c_p dT/T(V) - Rdp/p(V) and then integrated, I should not expect to get the same result as substituting T(V) and P(V) into the ΔS equation (as we had done earlier) as dS ≠ δQ/T?

If the system is the gas alone, then substituting the initial and final states into the ΔS equation gives the correct result, because, for the gas alone, δQ was not equal to zero. And, in our problem, with the system taken as the gas only, the expansion is reversible. And for that path for the gas, dS=δQ/T. However, if we took the system as the cylinder, piston, and gas, since this system is insulated, and δQ for that system = 0 for the entire path. As we've said earlier, it all depends on what you take as your system.

Since the piston ΔS = 0 anyways, does this mean that had I chosen to add F/A to P(V) (essentially use P_ext instead), I should still get the same change in entropy using the ΔS equation, or does that equation assume the entire system is an ideal gas?

That would assume that the entire system is an ideal gas, which it isn't. You can only use the ΔS equation we employed for an ideal gas.

Try to think of the entropy as a physical property of the material (in this case an ideal gas) rather than something associated with the specific nature of the process path.Chet

 

[Red_CCF]

Not really. P in the figure is what we have been calling P_ext, which inherently assumes that the gas, cylinder, and piston constitute the system. On expansion, P_ext is less than the gas pressure would have been if the gas expansion were reversible. (See if you can verify this.) On compression, P_ext is greater than the gas pressure would have been if the gas expansion were reversible. (See if you can verify this.)

Pext= P_o\left(\frac{V_o}{V}\right)^{γ}+\frac{FR}{AC_p}\left[1-\left(\frac{V_o}{V}\right)^γ\right] - F/A

Pext= P_o\left(\frac{V_o}{V}\right)^{γ}+\frac{F}{A}\left[\frac{R}{c_p}(1-\left(\frac{V_o}{V}\right)^γ)-1\right]

R/c_p = 1 - c_v/c_p &lt; 1
1-(\frac{V_o}{V})^γ &lt; 1

Hence the second term in the Pext equation is negative, and Pext is always lower than that from the reversible case (where Pext = P(V)). So the gas pressure/temperature (P(V), T(V)) always increases with friction regardless of expansion or compression, but the applied pressure to the surroundings is lower in expansion but higher in compression?

No, the ΔS equation applies to an ideal gas between two equilibrium states, irrespective of whether the actual process path was reversible or not.

I think I worded my question poorly. I meant to ask, although the final ΔS equation is valid for any process, my understanding is that the differential form dS = c_p dT/T - R dp/p was constructed for a reversible process (as it assumes TdS = Q) and integrated along a reversible path (due to the substitution of the ideal gas law)?

That would assume that the entire system is an ideal gas, which it isn't. You can only use the ΔS equation we employed for an ideal gas.

Try to think of the entropy as a physical property of the material (in this case an ideal gas) rather than something associated with the specific nature of the process path.

Would this be due to the fact that c_p comes into play (as it is a material property)?

Thanks

 

[Chestermiller]

Pext= P_o\left(\frac{V_o}{V}\right)^{γ}+\frac{FR}{AC_p}\left[1-\left(\frac{V_o}{V}\right)^γ\right] - F/A

Pext= P_o\left(\frac{V_o}{V}\right)^{γ}+\frac{F}{A}\left[\frac{R}{c_p}(1-\left(\frac{V_o}{V}\right)^γ)-1\right]

R/c_p = 1 - c_v/c_p &lt; 1
1-(\frac{V_o}{V})^γ &lt; 1

Hence the second term in the Pext equation is negative, and Pext is always lower than that from the reversible case (where Pext = P(V)). So the gas pressure/temperature (P(V), T(V)) always increases with friction regardless of expansion or compression, but the applied pressure to the surroundings is lower in expansion but higher in compression?

Yes.

I think I worded my question poorly. I meant to ask, although the final ΔS equation is valid for any process, my understanding is that the differential form dS = c_p dT/T - R dp/p was constructed for a reversible process (as it assumes TdS = Q) and integrated along a reversible path (due to the substitution of the ideal gas law)?

The only way I know of for getting the change in entropy between two equilibrium states of a system is to follow a reversible path between the two equilibrium states. The TdS equations provide an automatic way of doing this. But in our problem, we could just as easily have followed a path in which we first adiabatically and reversibly expanded the gas to its final volume, and then reversibly transferred heat at constant volume to arrive at the final temperature. Or we could have expanded it isothermally and reversibly to its final pressure and then transferred heat reversibly at constant pressure to bring it to its final temperature.

Would this be due to the fact that c_p comes into play (as it is a material property)?

Not really. The entropy per unit mass is a function of state of the material, every bit as much as internal energy, enthalpy, and heat capacity. For a single phase material, it is a function of temperature and pressure (or any other pair of intensive properties). Don't forget to consider the statistical thermo interpretation of entropy, which is related to the number of quantum mechanical states that the material exhibits.

Hey Red_CCF, remember when we were talking about the analogy between a gas undergoing an irreversible expansion (e.g., say free expansion) and the response of a multiple spring-mass system that is suddenly released from being compressed. Well the following link has an analysis and results for this problem (see post #16) in the limit of a continuous spring:
https://www.physicsforums.com/showthread.php?t=752813

Look it over, especially the results. Note the non-uniform nature of the spring compression with spatial position during the expansion.

Chet

 

[Red_CCF]

With regards to processes involving a massless piston, I am wondering, in these hypothetical scenarios, are these processes completed instantaneously? Since the piston is massless, it cannot accelerate, thus if beginning at its initial equilibrium state where P_{ext, i} = P_I, and suddenly dP is added (infinitessimal disturbance for quasistatic) or ΔP (finite disturbance for non-quasistatic) to P_{ext, i}, by Newton's third law the P_I must balance P_{ext, i} + dP or ΔP; does this mean that P_I should balance the disturbance at immediately as it occurs?

Not really. The entropy per unit mass is a function of state of the material, every bit as much as internal energy, enthalpy, and heat capacity. For a single phase material, it is a function of temperature and pressure (or any other pair of intensive properties).

I can kind of see the analogy between internal energy and entropy, but I'm having trouble interpreting it as material specific; what if the system is pure but is multiphased, or the system undergoes a chemical reaction (such that the "material" changes), or the system has components of different substances and phases (maybe solid metal with gas like we have), or even gases like air with multiple species; how does one generalize entropy to describe that of a whole system if the system itself has multiple components?

Thanks very much

 

[Chestermiller]

With regards to processes involving a massless piston, I am wondering, in these hypothetical scenarios, are these processes completed instantaneously? Since the piston is massless, it cannot accelerate, thus if beginning at its initial equilibrium state where P_{ext, i} = P_I, and suddenly dP is added (infinitessimal disturbance for quasistatic) or ΔP (finite disturbance for non-quasistatic) to P_{ext, i}, by Newton's third law the P_I must balance P_{ext, i} + dP or ΔP; does this mean that P_I should balance the disturbance at immediately as it occurs?

Sure. That's what Newton's second law tells us. Just draw a free body diagram of the piston. If the piston is massless, then ma = 0.

I can kind of see the analogy between internal energy and entropy, but I'm having trouble interpreting it as material specific; what if the system is pure but is multiphased, or the system undergoes a chemical reaction (such that the "material" changes), or the system has components of different substances and phases (maybe solid metal with gas like we have), or even gases like air with multiple species; how does one generalize entropy to describe that of a whole system if the system itself has multiple components?

If you have multiple phases of a single substance, then, no problem. You just get the entropy of the material in each phase, taking into account the change in entropy for the phase transition.

In the case of mixtures or solutions (chemical reaction, multiple components), J. Williard Gibbs figured out how to extend the theory to include this back in the early 1900's. For ideal gas mixtures, this is presented very nicely in Chapter 10 of Smith and Van Ness. What you end up doing is determining the entropy per unit mass of each individual species in the mixture. This requires the inclusion of an additional term in the equation for the specific entropy of each species (over and above the total pressure ratio term and the temperature ratio term).

Take a look at Chapter 10, and carefully read over the section on ideal gas mixtures. I think you'll find it very worthwhile.

Chet

 

[Red_CCF]

Sure. That's what Newton's second law tells us. Just draw a free body diagram of the piston. If the piston is massless, then ma = 0.

Would one be able to analyze the effects of viscous dissipations with a massless piston involved and/or would there be an error if this is done? If P_I (or σ_I) equilibrates to P_ext + dP or ΔP instantaneously,, wouldn't that have a significant effect on du/dx required to find τ_xx?

I was also thinking a little more about static/kinetic friction effects. If the piston is stationary before a compression process such that P_I + F_static/A = P_ext, once dP to get the piston moving, friction becomes kinetic and instantaneously drops such that there is a finite pressure difference between P_I + F_kinetic/A and P_ext + dP. If the piston is massless, P_I would increase instantaneously such that P_I + F_kinetic/A = P_ext + dP and since the piston is massless it should stop. But a finite pressure must be added to P_ext in addition to dP to overcome the static friction to get the piston moving again. Is the above still considered quasistatic, or is there some assumption we can toss into ignore this (and which friction would we end up using)?

In the case of mixtures or solutions (chemical reaction, multiple components), J. Williard Gibbs figured out how to extend the theory to include this back in the early 1900's. For ideal gas mixtures, this is presented very nicely in Chapter 10 of Smith and Van Ness. What you end up doing is determining the entropy per unit mass of each individual species in the mixture. This requires the inclusion of an additional term in the equation for the specific entropy of each species (over and above the total pressure ratio term and the temperature ratio term).

Take a look at Chapter 10, and carefully read over the section on ideal gas mixtures. I think you'll find it very worthwhile.

Chet

I think this came up in the combustion book I am reading; they essentially applied the ΔS equation we used and calculating ∑ΔS_i where ΔS_i is the entropy for each gas species?

Thanks very much

 

[Chestermiller]

Would one be able to analyze the effects of viscous dissipations with a massless piston involved and/or would there be an error if this is done? If P_I (or σ_I) equilibrates to P_ext + dP or ΔP instantaneously,, wouldn't that have a significant effect on du/dx required to find τ_xx?

Sure. If you were doing a gas dyanmics analysis (compressible fluid mechanics), this would be handled in the boundary condition at the piston face.

I was also thinking a little more about static/kinetic friction effects. If the piston is stationary before a compression process such that P_I + F_static/A = P_ext, once dP to get the piston moving, friction becomes kinetic and instantaneously drops such that there is a finite pressure difference between P_I + F_kinetic/A and P_ext + dP. If the piston is massless, P_I would increase instantaneously such that P_I + F_kinetic/A = P_ext + dP and since the piston is massless it should stop. But a finite pressure must be added to P_ext in addition to dP to overcome the static friction to get the piston moving again. Is the above still considered quasistatic, or is there some assumption we can toss into ignore this (and which friction would we end up using)?

The coefficient of kinetic friction is less than the coefficient of static friction, so, if F decreases, σ(I) increases. The gas near the interface would start compressing at just the right rate so that the force balance is still satisfied. Also, again, if the piston is massless, equality of the forces does not mean that the piston is not moving or accelerating. So the piston won't stop moving once the coefficient of static friction is exceeded. However, in this deformation, you can be certain that the deformation will not be quasistatic.

I think this came up in the combustion book I am reading; they essentially applied the ΔS equation we used and calculating ∑ΔS_i where ΔS_i is the entropy for each gas species?

For this to be correct, they would have to use the partial pressure of the individual gas species in calculating the ΔS_i, rather than the total pressure of the combination of gases. Is this what they did?

Chet

 

[Red_CCF]

Sure. If you were doing a gas dyanmics analysis (compressible fluid mechanics), this would be handled in the boundary condition at the piston face.

What is the boundary condition at the piston-gas interface for a massless piston? Does the piston at any time have a measurable velocity if it cannot accelerate and equilibrium is instantaneous?

The coefficient of kinetic friction is less than the coefficient of static friction, so, if F decreases, σ(I) increases. The gas near the interface would start compressing at just the right rate so that the force balance is still satisfied. Also, again, if the piston is massless, equality of the forces does not mean that the piston is not moving or accelerating. So the piston won't stop moving once the coefficient of static friction is exceeded. However, in this deformation, you can be certain that the deformation will not be quasistatic.

So the piston will move Δx (instead of dx due to the finite pressure difference) instantaneously such that σ(I) can increase instantaneously as F changes from static to kinetic in order to satisfy Newton's Second Law for a massless piston? I assume that the pressure of the rest of the system cannot react as fast, thus the experiment we were analyzing even theoretically must be nonquasistatic?

Since the piston has no inertia, I assume that once σ(I) and kinetic friction balances the external forces the piston must stop, friction becomes static again, and the cycle continues?

For this to be correct, they would have to use the partial pressure of the individual gas species in calculating the ΔS_i, rather than the total pressure of the combination of gases. Is this what they did?

Chet

The exact equation for the entropy of the ith species of the combustion products mixture given was:

s_i = s_i^o (Tref) + ∫c_p\frac{dT}{T} - R_uln(\frac{P_i}{P^o})

where the integration was taken from some reference state T_ref to T_f; p_i is the species partial pressure.

Although it is not stated, I am assuming that total entropy change of the reaction is ∑s_i,products - ∑s_i,reactants.

Is the point of having a reference state just so that the lower limit of the integral is known and we only need to add s_i^o(T_ref) to find absolute entropy? Is there an issue of integrating from the state where s = 0 and if so how were the tabulated s values measured?

Thanks

 

[Chestermiller]

What is the boundary condition at the piston-gas interface for a massless piston?

Irrespective of whether the piston is massless, the boundary condition at the piston-gas interface (using BSL notation, in which compressive stresses are positive) is

F=σ_IA=(p-2μ\frac{∂v}{∂x})A

where F is the force that the gas exerts on the piston face, A is the area of the piston, p is the local thermodynamic pressure of the gas at the interface =ρRT, ρ is the local molar gas density at the interface, μ is the gas viscosity at the interface, and ∂v/∂x = the spatial axial velocity gradient of the gas in the vicinity of the interface.

Does the piston at any time have a measurable velocity if it cannot accelerate and equilibrium is instantaneous?

Who says the piston can't accelerate. Suppose the mass of the piston is 10^-15 grams. How much of a force difference between the two faces of the piston does it take in order to accelerate it?

So the piston will move Δx (instead of dx due to the finite pressure difference) instantaneously such that σ(I) can increase instantaneously as F changes from static to kinetic in order to satisfy Newton's Second Law for a massless piston?

No. The piston doesn't move Δx instantaneously. All that is required for σ(I) to increase. This will happen as soon as the piston starts moving. The viscous stress contribution causes σ(I) to increase. Even more importantly, the local gas density in close proximity to the interface will increase, resulting in a local increase in the thermodynamic pressure p. This is because, as the piston begins to move, the entirety of the gas cannot compress all at once because of its inertia. So a small region of higher density begins to develop adjacent to the piston face. This higher density region will grow in extent as time progresses.

I assume that the pressure of the rest of the system cannot react as fast, thus the experiment we were analyzing even theoretically must be nonquasistatic?

Correct.

Since the piston has no inertia, I assume that once σ(I) and kinetic friction balances the external forces the piston must stop, friction becomes static again, and the cycle continues?

No. As we said, the acceleration of the piston is not zero. Consider again the case where the mass of the piston is 10^-15 grams, rather than zero. The difference in forces on the two sides of the piston necessary to accelerate the piston and keep it movingwill be very close to zero.

The exact equation for the entropy of the ith species of the combustion products mixture given was:

s_i = s_i^o (Tref) + ∫c_p\frac{dT}{T} - R_uln(\frac{P_i}{P^o})

where the integration was taken from some reference state T_ref to T_f; p_i is the species partial pressure.

Although it is not stated, I am assuming that total entropy change of the reaction is ∑s_i,products - ∑s_i,reactants.

No. The entropies need to be weighted in terms of the mole fractions.

Is the point of having a reference state just so that the lower limit of the integral is known and we only need to add s_i^o(T_ref) to find absolute entropy? Is there an issue of integrating from the state where s = 0 and if so how were the tabulated s values measured?

No. The absolute entropy is never used. Everything is referenced to the entropy of each pure species in the reference state. The tabulated values of the pure species entropies in the reference state are determined from the free energies of formation and heats of formation of the pure species in the reference state. The free energies of formation are "backed out" from measurements of the equilibrium constants for various equilibrium reactions, and the heats of formation are "backed out" from the heats of reaction for these reactions. Once the free energies of formation and heats of formation of the species are established, these values can be used to determine the equilibrium constants and heats of reaction for any other reactions involving species included in the overall list of species. This is what makes the methodology so powerful.

Chet

 

[Red_CCF]

No. The piston doesn't move Δx instantaneously. All that is required for σ(I) to increase. This will happen as soon as the piston starts moving. The viscous stress contribution causes σ(I) to increase. Even more importantly, the local gas density in close proximity to the interface will increase, resulting in a local increase in the thermodynamic pressure p. This is because, as the piston begins to move, the entirety of the gas cannot compress all at once because of its inertia. So a small region of higher density begins to develop adjacent to the piston face. This higher density region will grow in extent as time progresses.

If the process was truly quasistatic, the system pressure = interface pressure = ideal gas pressure and thus pressure change is dependent on the volume change during compression. In reality, does the change in p(I) have little dependence on the volume change of the system and thus no longer quasistatic? Does σ(I) immediately compensate for the drop in friction from static to kinetic (σ(I) + F_kin/A = P_ext +dP) such that the net force on the massless piston is zero?

Who says the piston can't accelerate. Suppose the mass of the piston is 10^-15 grams. How much of a force difference between the two faces of the piston does it take in order to accelerate it?

No. As we said, the acceleration of the piston is not zero. Consider again the case where the mass of the piston is 10^-15 grams, rather than zero. The difference in forces on the two sides of the piston necessary to accelerate the piston and keep it movingwill be very close to zero.

My thought is that since net force must equal 0 per Newton's Second Law, essentially 0 = 0*a, but now a is undefined since there's an infinite number of solutions, so how do we know the kinematics of the piston (displacement, velocity, acceleration) for every dP addition?

No. The absolute entropy is never used. Everything is referenced to the entropy of each pure species in the reference state. The tabulated values of the pure species entropies in the reference state are determined from the free energies of formation and heats of formation of the pure species in the reference state. The free energies of formation are "backed out" from measurements of the equilibrium constants for various equilibrium reactions, and the heats of formation are "backed out" from the heats of reaction for these reactions. Once the free energies of formation and heats of formation of the species are established, these values can be used to determine the equilibrium constants and heats of reaction for any other reactions involving species included in the overall list of species. This is what makes the methodology so powerful.

Chet

Is absolute entropy used when finding ΔS (pre and post combustion), as the equation in my book seem to imply this by having the reference entropy on the right side?

Is the reference state entropy is analogous/equivalent to the enthalpy of formation component of a specie's absolute enthalpy? How come it is independent pressure unlike the enthalpy of formation?

Thank you very much

 

[Chestermiller]

If the process was truly quasistatic, the system pressure = interface pressure = ideal gas pressure and thus pressure change is dependent on the volume change during compression. In reality, does the change in p(I) have little dependence on the volume change of the system and thus no longer quasistatic?

What happens in irreversible non-quasistatic compression is that gas immediately adjacent to the piston becomes compressed first, while the gas further away hasn't gotten compressed yet. This is what happens at short times. As time progresses, the compressed region of gas grows in spatial extent.

Does σ(I) immediately compensate for the drop in friction from static to kinetic (σ(I) + F_kin/A = P_ext +dP) such that the net force on the massless piston is zero?

Yes.

My thought is that since net force must equal 0 per Newton's Second Law, essentially 0 = 0*a, but now a is undefined since there's an infinite number of solutions, so how do we know the kinematics of the piston (displacement, velocity, acceleration) for every dP addition?

This is not correct. It should be F_net = 0*a = 0. Regarding the question about "how do we know the kinematics", the gas has mass/inertia, and you are using the piston to apply a force to it. To find out the time-dependent kinematics of the piston, you would have to solve the gas dynamics partial differential equations using Pext + dP as the time dependent stress boundary condition. Sometimes, this is what you would have to do to solve your problem, even if you are interested only in the initial and final equilibrium states.

Is absolute entropy used when finding ΔS (pre and post combustion), as the equation in my book seem to imply this by having the reference entropy on the right side?

I would have to see what your book has before I comment on this.

Is the reference state entropy is analogous/equivalent to the enthalpy of formation component of a specie's absolute enthalpy?

Yes. The enthalpy is usually referred to a reference state, rather than absolute zero.

How come it is independent pressure unlike the enthalpy of formation?

Neither the enthalpy of formation nor the entropy of formation is dependent on pressure, because they both refer to the pure specie at 1 atm.

Chet

 

[rkmurtyp]

If the process was truly quasistatic, the system pressure = interface pressure = ideal gas pressure and thus pressure change is dependent on the volume change during compression. In reality, does the change in p(I) have little dependence on the volume change of the system and thus no longer quasistatic? Does σ(I) immediately compensate for the drop in friction from static to kinetic (σ(I) + F_kin/A = P_ext +dP) such that the net force on the massless piston is zero?

Since you are also referring to chemical reactions as examples of processes under discussion, the following might help.

In the study of chemical reactions, besides thermodynamic considerations, we study kinetics of reactions to arrive at mechanism (Path followed by the reaction in going from a stste A to a state B) of reaction. Unlike the definite (unique) values for changes in the thermodynamic properties, the values of kinetic properties such as rate constants, do not have unique values. We can think of a number of (possible) mechanisms each of which satisfies some of the experimentally observed/measured properties for a given reaction.

In a similar fashion we can give a number of alternate explanations (for the rate of motion of pistion in the presence/absence of friction, whether we call it quasistatic process or irreversible process etc.

In short, uiqueness is lost in rates of processes - plurality of paths connecting two given states becomes possible, which is not the case with thermodynamics of the process. Thermodynamics gives 'yes' or 'no' type of answers ( a process is either reversible or irreversble - no plurality of answers!) with 100% certainity.

In essence, even if we arrive at a possible explanation of how the piston (whether massless or otherwise) moves, with some model for viscocity, pressure distribution etc of the gas, it will not be unique - there exists no way we can prove that it is the correct or unique path the process followed.

rkmurty

 

[Chestermiller]

In short, uiqueness is lost in rates of processes - plurality of paths connecting two given states becomes possible, which is not the case with thermodynamics of the process. Thermodynamics gives 'yes' or 'no' type of answers ( a process is either reversible or irreversble - no plurality of answers!) with 100% certainity.

In essence, even if we arrive at a possible explanation of how the piston (whether massless or otherwise) moves, with some model for viscocity, pressure distribution etc of the gas, it will not be unique - there exists no way we can prove that it is the correct or unique path the process followed.

If what you're saying here were correct, then we Chemical Engineers could never have done what we have been doing successfully for over a hundred years, which is confidently design and operate real world chemical plants involving equipment such as compressors, heat exchangers, distillation towers, absorption columns, chemical reactors, cooling towers, dryers, adsorption beds, ion exchange columns, evaporators, piping networks, filters, etc.

Chet

 

[Red_CCF]

What happens in irreversible non-quasistatic compression is that gas immediately adjacent to the piston becomes compressed first, while the gas further away hasn't gotten compressed yet. This is what happens at short times. As time progresses, the compressed region of gas grows in spatial extent.

In the example we were solving involving wall friction, in reality stress from gas viscosity must always be significant enough to render the process non-quasistatic to compensate for the static-kinetic friction change? Would this process approach quasistatic if wall frictional effects approaches 0?

This is not correct. It should be F_net = 0*a = 0. Regarding the question about "how do we know the kinematics", the gas has mass/inertia, and you are using the piston to apply a force to it. To find out the time-dependent kinematics of the piston, you would have to solve the gas dynamics partial differential equations using Pext + dP as the time dependent stress boundary condition. Sometimes, this is what you would have to do to solve your problem, even if you are interested only in the initial and final equilibrium states.

So a massless piston's acceleration can be any number including 0 per the gas dynamic equations?

Does the piston stop once σ(I) + Fkin/A = Pext +dP? If the piston continues to move, then σ(I) will continue to increase, and I can't wrap my head around what happens next.

I would have to see what your book has before I comment on this.

Yes. The enthalpy is usually referred to a reference state, rather than absolute zero.

Neither the enthalpy of formation nor the entropy of formation is dependent on pressure, because they both refer to the pure specie at 1 atm.

Chet

Unfortunately the book I have doesn't say much about the second law, just stuck in two equations in an example that computes product entropy (species and mixture) attached. The plot given was w.r.t. S of the mixture. It seems most books just state the enthalpy/entropy fundamental equations as facts without much explanation.

If both enthalpy/entropy of formation are independent of pressure, does this mean that if temperature is kept constant, the enthalpy/entropy of formation of any substance is constant w.r.t pressure (i.e. forming CO2 at 1atm or 100atm releases the same amount of energy)?

Thanks very much

 

[rkmurtyp]

If what you're saying here were correct, then we Chemical Engineers could never have done what we have been doing successfully for over a hundred years, which is confidently design and operate real world chemical plants involving equipment such as compressors, heat exchangers, distillation towers, absorption columns, chemical reactors, cooling towers, dryers, adsorption beds, ion exchange columns, evaporators, piping networks, filters, etc.

Chet

No doubt, we have been designing and operating equipment and processes in everyday practice. However, for each of those designs, we can have several alternate possible designs, the design takes several other factors (such as financial aspects) into consideration. Suppose we ask if a given design for a process is unique - do we get the answer, yes or no? We don't.

In contrast, thermodynamics of a process yeilds unique results/answers. For example, for questions such as: Can the system reach state B spontaneously from state A? we have unique answers, whereas, with regard to rate processes we can not get such unique answers.

rkmurty

 

[Chestermiller]

No doubt, we have been designing and operating equipment and processes in everyday practice. However, for each of those designs, we can have several alternate possible designs, the design takes several other factors (such as financial aspects) into consideration. Suppose we ask if a given design for a process is unique - do we get the answer, yes or no? We don't.

With all due respect, I totally disagree. If I design any of the pieces of equipment I mentioned in my previous post, I fully expect the model predictions to match the observed performance of the equipment with respect to temperatures, pressures, species concentrations, etc. entering and exiting the equipment, as well as throughout the equipment. This kind of predictive capability has validated observationally time and time again. If we engineers couldn't do these things, no one would be shelling out multi-billions of dollars to build chemical plants, and no one would be paying us engineers to big buck to design them.

Having an "all bets are off attitude" like this is highly detrimental. Before you go around telling us engineers what we are able and not able to do, maybe you should study engineering yourself and ascertain the true facts.
[/QUOTE]

 

[rkmurtyp]

With all due respect, I totally disagree. If I design any of the pieces of equipment I mentioned in my previous post, I fully expect the model predictions to match the observed performance of the equipment with respect to temperatures, pressures, species concentrations, etc. entering and exiting the equipment, as well as throughout the equipment. This kind of predictive capability has validated observationally time and time again. If we engineers couldn't do these things, no one would be shelling out multi-billions of dollars to build chemical plants, and no one would be paying us engineers to big buck to design them.

Having an "all bets are off attitude" like this is highly detrimental. Before you go around telling us engineers what we are able and not able to do, maybe you should study engineering yourself and ascertain the true facts.

[/QUOTE]

Thanks

 

[Red_CCF]

With all due respect, I totally disagree. If I design any of the pieces of equipment I mentioned in my previous post, I fully expect the model predictions to match the observed performance of the equipment with respect to temperatures, pressures, species concentrations, etc. entering and exiting the equipment, as well as throughout the equipment. This kind of predictive capability has validated observationally time and time again. If we engineers couldn't do these things, no one would be shelling out multi-billions of dollars to build chemical plants, and no one would be paying us engineers to big buck to design them.

Having an "all bets are off attitude" like this is highly detrimental. Before you go around telling us engineers what we are able and not able to do, maybe you should study engineering yourself and ascertain the true facts.

Hi Chet

When you get a chance, can you take a look at some of the questions I had in post #54 here

Thanks!

 

[Chestermiller]

Hi Chet

When you get a chance, can you take a look at some of the questions I had in post #54 here

Thanks!

Hi Red_CCF. Rest assured that I haven't forgotten #54. I just wanted to take a little more time than usual thinking about exactly how I wanted to answer these questions in an effective way.

Chet

 

[Chestermiller]

In the example we were solving involving wall friction, in reality stress from gas viscosity must always be significant enough to render the process non-quasistatic to compensate for the static-kinetic friction change?

The main contributor to the stress is the increased local pressure in the portion of the gas immediately adjacent to the piston. The viscous stress is secondary, but important long term. As the piston advances, the extent of the compressed gas region increases, and the newly accelerated gas allows the pressure at the piston face to be maintained. All this is happening at short times.

Would this process approach quasistatic if wall frictional effects approaches 0?
So a massless piston's acceleration can be any number including 0 per the gas dynamic equations? Does the piston stop once σ(I) + Fkin/A = Pext +dP? If the piston continues to move, then σ(I) will continue to increase, and I can't wrap my head around what happens next.

The answers to these questions depend on how we control the motion of the piston. We have total control on the external pressure P_ext(t) that we apply and/or the kinematics of the piston motion. Imagine that there is a push-rod attached to the top of the piston, and we control the motion of the rod by hand. When we feel the static friction give way, we can back off on the pressure we apply, so that the piston moves at whatever slow velocity we wish. Or we can try to hold the force we apply constant at the value that existed when the static friction gave way, in which case the gas immediately adjacent to the piston will start to accelerate. We can apply any motion we desire to the piston by controlling P_ext(t). When we do this, the non-uniform deformation within the gas will adjust itself in such a manner that σ(I) always satisfies the equation σ(I) + Fkin/A = Pext(t).

Unfortunately the book I have doesn't say much about the second law, just stuck in two equations in an example that computes product entropy (species and mixture) attached. The plot given was w.r.t. S of the mixture. It seems most books just state the enthalpy/entropy fundamental equations as facts without much explanation.

Sorry your book doesn't do a good job. I looked over the excerpt you sent, and it (at least) looks correct. What they are trying to do is calculate what the entropy of the reaction mixture would be at various conversions. The conversion that maximizes the entropy for this isolated reacting system is the one that corresponds to the equilibrium conversion.

If both enthalpy/entropy of formation are independent of pressure, does this mean that if temperature is kept constant, the enthalpy/entropy of formation of any substance is constant w.r.t pressure (i.e. forming CO2 at 1atm or 100atm releases the same amount of energy)?

It's not quite correct to say that the enthalpy/entropy of formation are "independent of pressure." Both the enthalpy of formation and the entropy of formation occur at a constant pressure of one atmosphere. For an ideal gas, the enthalpy is independent of pressure, but the entropy definitely depends on a pressure. After the material is formed from the elements at 1 atm. pressure, if the pressure of the species changes, its entropy changes.

Chet