Question about Flow between Parallel Plates

[Chestermiller]

With regards to h^o(T), per the first quote, isn't that the same as the enthalpy defined relative to standard state?

Yes, provided what you call the standard state refers its elements at the standard state of 298 and 1 atm.

For the second quote, my interpretation is that h^o(T) is defined relative to the the stoichiometric proportions of its elements at its actual T (so I need to subtract the sensible heat of the elements from 298K to T of the compound) as opposed to just at 298K (per standard state definition), is this correct?

I was unable to find the equation I was referring to in the quote. But, in any event, your statement is incorrect. h^o(T) is defined at temperature T relative to the the stoichiometric proportions of its elements at 298 and 1 atm (the standard reference state). On the other hand, h_f° is defined at temperature T relative to the stoichiometric proportions of its elements at its actual T.

This implies h°(T)=H°(T)-∑n_iH°_i(T) which for elements is 0 for all T which didn't make much sense to me.

No. This is the definition of h_f°(T).

Is h°(T)=H°(T)-∑n_iH°_i(T)+∑n_ih°_i(T) (based on the equality in h_f°(T))?

This is incorrect.

For H°(T)-∑n_iH_i°(T)=h°(T)-∑n_ih_i°(T), per definition of H°(T) and h°(T), the sensible components cancel as

H°(T)-h°(T)=∑n_iH_i°(T)-∑n_ih_i°(T)

and the equation reduces to H°(298)-∑n_iH_i°(298)=h°(298)-∑n_ih_i°(298) = h°(298). What I'm left with is that h_f°(T) is constant for all T?

This is not done correctly. And, if done correctly, the sensible heats do not cancel.

Let's start with the equation I gave for h_f°(T):

h_f°(T)=H°(T)-∑n_iH°_i(T)

Now, $H°(T)=H°(298)+\int_{298}^T{C_pdT}$
and, for its elements, $H_i°(T)=H_i°(298)+\int_{298}^T{C_{pi}dT}$
If we substitute this into the equation for h_f°(T), we obtain:

$$h_f°(T)=(H°(298)-∑n_iH_i°(298))+\int_{298}^T{C_pdT}-∑n_i\int_{298}^T{C_{pi}dT}$$
But, $(H°(298)-∑n_iH_i°(298))=h°(298)$
So,
$$h_f°(T)=h°(298)+\int_{298}^T{C_pdT}-∑n_i\int_{298}^T{C_{pi}dT}$$
But, $h°(298)+\int_{298}^T{C_pdT}=h°(T)$
and $0+\int_{298}^T{C_{pi}dT}=h_i°(T)$ for the elements.
So,
$$h_f°(T)=h°(T)-∑n_ih_i°(T)$$

Chet

 

[Red_CCF]

Yes, provided what you call the standard state refers its elements at the standard state of 298 and 1 atm.

Let's start with the equation I gave for h_f°(T):

h_f°(T)=H°(T)-∑n_iH°_i(T)

Now, $H°(T)=H°(298)+\int_{298}^T{C_pdT}$
and, for its elements, $H_i°(T)=H_i°(298)+\int_{298}^T{C_{pi}dT}$
If we substitute this into the equation for h_f°(T), we obtain:

$$h_f°(T)=(H°(298)-∑n_iH_i°(298))+\int_{298}^T{C_pdT}-∑n_i\int_{298}^T{C_{pi}dT}$$
But, $(H°(298)-∑n_iH_i°(298))=h°(298)$
So,
$$h_f°(T)=h°(298)+\int_{298}^T{C_pdT}-∑n_i\int_{298}^T{C_{pi}dT}$$
But, $h°(298)+\int_{298}^T{C_pdT}=h°(T)$
and $0+\int_{298}^T{C_{pi}dT}=h_i°(T)$ for the elements.
So,
$$h_f°(T)=h°(T)-∑n_ih_i°(T)$$

Per this definition then is h°(T) = H°(T) - ∑n_iH_i(298)? If so, then is the following correct to show the relationship you provided for h°(T):

$h°(T) = H°(T) - ∑n_iH_i(298) = H°(298) - ∑n_iH_i(298) + \int_{298}^T{C_{pi}dT}$
$h°(T) = h°(298) + \int_{298}^T{C_{pi}dT}$

Can I interpret this as equivalent to the standard enthalpy conventionally defined and used in textbooks unless there are some minor differences that I am not seeing?

Essentially the process we went through was, define an absolute enthalpy H°(T) relative to absolute zero, then the enthalpy of formation becomes h_f°(T) = H°(T) - ∑n_iH_i(T). This is followed by defining the relative enthalpy h°(T) = H°(T) - ∑n_iH_i(298) that became the standard state enthalpy (if my first line is correct) and expressing h_f°(T) in terms of h°(T) and h°(T) in terms of h°(298) follow from the above definitions, is this correct?

Thanks very much

 

[Chestermiller]

Per this definition then is h°(T) = H°(T) - ∑n_iH_i(298)?

Yes. Yes. and Yes.

If so, then is the following correct to show the relationship you provided for h°(T):

$h°(T) = H°(T) - ∑n_iH_i(298) = H°(298) - ∑n_iH_i(298) + \int_{298}^T{C_{pi}dT}$
$h°(T) = h°(298) + \int_{298}^T{C_{pi}dT}$

Yes, if you remove the "i" subscript from C_p, since, in what we have been doing, we have been using the i subscript to refer to the elements.

Can I interpret this as equivalent to the standard enthalpy conventionally defined and used in textbooks unless there are some minor differences that I am not seeing?

I think so.

Essentially the process we went through was, define an absolute enthalpy H°(T) relative to absolute zero, then the enthalpy of formation becomes h_f°(T) = H°(T) - ∑n_iH_i(T). This is followed by defining the relative enthalpy h°(T) = H°(T) - ∑n_iH_i(298) that became the standard state enthalpy (if my first line is correct) and expressing h_f°(T) in terms of h°(T) and h°(T) in terms of h°(298) follow from the above definitions, is this correct?

Yes. I think you finally have it. But please understand that, in typical developments, they don't find it necessary to introduce the absolute enthalpy (whatever the absolute enthalpy means) into the analysis. I only introduced this formalism here because I thought it might make it easier for you to overcome your barrier to understanding. Of course, in the case of entropy, the absolute entropy is often used, and the development exactly parallels the one you described above (except, of course, for using C_p/T in place of C_p).

Chet

 

[Red_CCF]

Yes. I think you finally have it. But please understand that, in typical developments, they don't find it necessary to introduce the absolute enthalpy (whatever the absolute enthalpy means) into the analysis. I only introduced this formalism here because I thought it might make it easier for you to overcome your barrier to understanding. Of course, in the case of entropy, the absolute entropy is often used, and the development exactly parallels the one you described above (except, of course, for using C_p/T in place of C_p).

Chet

Is there a reason why absolute entropy is used but absolute enthalpy/free energy is not, especially since I imagine that for free energy the entropy used is relative to standard state? Is it typical that enthalpy tables (like steam tables) use standard state enthalpy, as I compared the steam table in my thermodynamics book with H₂O in my combustion book and they are very different?

With regards to elements, are we defining it as the lowest energy state/most stable form we encounter so for something like H₂, the formation reaction would be H₂ -> H₂ (not something like 2H -> H₂)? For the formation of H i assume it is 1/2H₂ -> H?

Thanks very much

 

[Chestermiller]

Is there a reason why absolute entropy is used but absolute enthalpy/free energy is not, especially since I imagine that for free energy the entropy used is relative to standard state?

I don't know. I have never seen a single book that presents absolute enthalpy or absolute free energy, and I'm not even sure that such concepts make sense or can be determined. The practicality of the matter is that, to do industrial calculations (e.g., chemical equilibrium), we don't need to know these functions down to absolute zero, but we do need to know them at typical process temperatures, which are above 0C. So why bother to measure and determine these functions all the way down to absolute zero, when our methodology works perfectly well using the function values relative to the elements (in their natural states) at the reference conditions.

In many textbooks, they don't even work with entropy in determining equilibrium constants. They just give the enthalpy of formation and the free energy of formation at the reference state, together with the heat capacity as a function of temperature. This information is sufficient to determine the free energy of formation as a function of temperature at 1 atm., which is sufficient to determine the equilibrium constant as a function of temperature.

Is it typical that enthalpy tables (like steam tables) use standard state enthalpy, as I compared the steam table in my thermodynamics book with H₂O in my combustion book and they are very different?

Yes. For steam calculations, it is convenient to choose as the reference state liquid water at 0C and 1 atm. For chemical equilibrium calculations, the reference state is conventionally chosen as 25C and 1 atm. No big deal. When doing calculations, you just need to understand what reference state they are using.

With regards to elements, are we defining it as the lowest energy state/most stable form we encounter so for something like H₂, the formation reaction would be H₂ -> H₂ (not something like 2H -> H₂)?

Yes. It is the natural state at 25C and 1 atm.

For the formation of H i assume it is 1/2H₂ -> H?

Yes.

 

[Red_CCF]

I don't know. I have never seen a single book that presents absolute enthalpy or absolute free energy, and I'm not even sure that such concepts make sense or can be determined. The practicality of the matter is that, to do industrial calculations (e.g., chemical equilibrium), we don't need to know these functions down to absolute zero, but we do need to know them at typical process temperatures, which are above 0C. So why bother to measure and determine these functions all the way down to absolute zero, when our methodology works perfectly well using the function values relative to the elements (in their natural states) at the reference conditions.

In many textbooks, they don't even work with entropy in determining equilibrium constants. They just give the enthalpy of formation and the free energy of formation at the reference state, together with the heat capacity as a function of temperature. This information is sufficient to determine the free energy of formation as a function of temperature at 1 atm., which is sufficient to determine the equilibrium constant as a function of temperature.

Are there anything entropy is useful for in practice and is there a reason they are not used very much?

So both absolute and standard state references similar in the sense that species' enthalpy are defined relative to their constituent elements and only dissimilar in that the actual state of reference is different?

Just for theoretical purposes, not that it is useful, but if I wanted to find the relationship in enthalpy change from standard state to absolute zero, does one exist (something like ∫c_pdT but accounting for pressure change) or is this something that has to be measured empirically?

No, but we are just trying to address some the doubts you had regarding the transition from small finite piston mass to the limit of zero piston mass, such as:

1. How can you have an accelerating piston if the piston mass is zero?
2. Does the piston oscillate if its mass is very small, but not zero?
3. Is the transition smooth from small finite piston mass to zero piston mass?
This solution answers all these questions.

We can continue the solution to larger mass, if you'd like. If mk/C^2 is equal to 1, the system is critically damped. If it is greater than 1, then the system may oscillate, depending on whether static friction is re-established at the moment that the piston reaches its maximum compression (for the first time).

With regards to the piston example we did earlier, one key assumption was that ΔP_ext = 0 at all times. If this were not true (say for a quasistatic or a less "violent" non-quasistatic process), how would one approach this problem; is it possible without doing this numerically?

Thank you very much

 

[Chestermiller]

Are there anything entropy is useful for in practice and is there a reason they are not used very much?

Entropy and internal energy are fundamental to the whole development. The other thermodynamic functions, H, G, and A are derived from them. However, G and A are useful with regard to solving problems involving thermodynamic equilibrium (phase equilibrium and chemical equilibrium).

So both absolute and standard state references similar in the sense that species' enthalpy are defined relative to their constituent elements and only dissimilar in that the actual state of reference is different?

Yes.

Just for theoretical purposes, not that it is useful, but if I wanted to find the relationship in enthalpy change from standard state to absolute zero, does one exist (something like ∫c_pdT but accounting for pressure change) or is this something that has to be measured empirically?

Yes. Are you aware of the relationship between dH, and dT and dP for a single phase pure substance? Also, there may be some phase changes along the way.

With regards to the piston example we did earlier, one key assumption was that ΔP_ext = 0 at all times. If this were not true (say for a quasistatic or a less "violent" non-quasistatic process), how would one approach this problem; is it possible without doing this numerically?

If you specified what ΔP_ext was as a function of time, then you could solve for the corresponding piston displacement analytically. A numerical solution would not be required because, in our simplified model, the system is linear.

Chet

 

[Red_CCF]

Entropy and internal energy are fundamental to the whole development. The other thermodynamic functions, H, G, and A are derived from them. However, G and A are useful with regard to solving problems involving thermodynamic equilibrium (phase equilibrium and chemical equilibrium).

How are entropy and internal energy measured empirically as tabulated in textbooks?

Yes. Are you aware of the relationship between dH, and dT and dP for a single phase pure substance? Also, there may be some phase changes along the way.

Is it the following relationship:

$$dh = (\frac{\partial{h}}{\partial{T}})_PdT+(\frac{\partial{h}}{\partial{P}})_TdP$$

assuming integrating this is possible, the absolute enthalpy H(T, P) of a compound or element would be the integral of the first term on the RHS from 0K to T and the second term from 0atm to P and for H(T, 1 atm) = H^o(T) the pressure integral is always from 0 to 1 atm? I imagine that phase changes would be accounted for by separating integral into three parts and adding h_sublimation and h_vaporization (assuming we are ending up with a gas).

If you specified what ΔP_ext was as a function of time, then you could solve for the corresponding piston displacement analytically. A numerical solution would not be required because, in our simplified model, the system is linear.

Chet

If the process was dictated to be quasistatic, is it possible to predict a ΔP_ext(t) which will allow such process to occur, or must ΔP_ext be determined empirically? Also, does the addition of this term only change the particular/steady state solution of the ODE?

Thank you very much

 

[Chestermiller]

How are entropy and internal energy measured empirically as tabulated in textbooks?
You already did many problems about this when you studied the first and second laws. You just have to conceive of experiments that focus on these parameters. In the case of entropy, your experiment has to be done under nearly reversible conditions.

Is it the following relationship:

$$dh = (\frac{\partial{h}}{\partial{T}})_PdT+(\frac{\partial{h}}{\partial{P}})_TdP$$

Look this up in Smith and Van Ness. The expression for the partial of h with respect to P is obtained using a Maxwell relationship, and this derivative can be calculated from the P-V-T behavior of the material.

assuming integrating this is possible, the absolute enthalpy H(T, P) of a compound or element would be the integral of the first term on the RHS from 0K to T and the second term from 0atm to P and for H(T, 1 atm) = H^o(T) the pressure integral is always from 0 to 1 atm?

As I said, the second term is calculated from the P-V-T behavior of the material.

I imagine that phase changes would be accounted for by separating integral into three parts and adding h_sublimation and h_vaporization (assuming we are ending up with a gas).

The integrations would be done backwards, starting from the ideal gas state at normal temperatures and low pressures. For the ideal gas region, h is not a function of pressure.

If the process was dictated to be quasistatic, is it possible to predict a ΔP_ext(t) which will allow such process to occur, or must ΔP_ext be determined empirically?

We already did quasistatic cases in which the acceleration and the velocity approach zero. You yourself discussed a case where you apply a short initial small acceleration, followed by a constant small velocity, followed by a short final deceleration. I discussed a case of a very small acceleration applied over the entire deformation, followed by a short deceleration to stop the deformation. For both these cases, ΔP_ext(t) is slightly different, but the difference is insignificant in the quasistatic limit. Both these cases give the exact same result for ΔP_ext as a function of Δx in the limit in which the acceleration and the velocity approach zero. This is the quasistatic case.

Also, does the addition of this term only change the particular/steady state solution of the ODE?

No. It changes the entire history of the deformation. However, in the limit of quasistatic deformation, every point along the transition is essentially steady state.

Chet

 

[Red_CCF]

Look this up in Smith and Van Ness. The expression for the partial of h with respect to P is obtained using a Maxwell relationship, and this derivative can be calculated from the P-V-T behavior of the material.

As I said, the second term is calculated from the P-V-T behavior of the material.

The integrations would be done backwards, starting from the ideal gas state at normal temperatures and low pressures. For the ideal gas region, h is not a function of pressure.

Were you referring to the following relationship for ∂h/∂P:

$$\frac{∂h}{∂P} = [V-T(\frac{∂V}{∂T})_P]dP$$

Does the above only applied to fluids (the book appear to imply this) or any phased material?

Realistically, what is the "path" that is acceptable to perform the integral? I assume that we cannot just integrate from 1 to 0 atm at constant 298K and then from 298K to 0K at constant 0 atm or vice versa? Typically at 1 atm, how low must the temperature go before we begin to deviate from ideal gas?

We already did quasistatic cases in which the acceleration and the velocity approach zero. You yourself discussed a case where you apply a short initial small acceleration, followed by a constant small velocity, followed by a short final deceleration. I discussed a case of a very small acceleration applied over the entire deformation, followed by a short deceleration to stop the deformation. For both these cases, ΔP_ext(t) is slightly different, but the difference is insignificant in the quasistatic limit. Both these cases give the exact same result for ΔP_ext as a function of Δx in the limit in which the acceleration and the velocity approach zero. This is the quasistatic case.

For quasistatic compression, many textbooks describes it generally as increasing external pressure by dP -> piston compresses -> wait until equilibrium is established -> repeat. By this logic I feel that the piston must always be accelerating since there is always a net external dP acting on the piston, is this correct? I had a feeling after further thought that a constant velocity quasistatic compression is not achievable since there is always a need for continuous net force?

No. It changes the entire history of the deformation. However, in the limit of quasistatic deformation, every point along the transition is essentially steady state.

Chet

With regards to the original equation copied below:

$$m\frac{d^2(Δx_1)}{dt^2}+C\frac{d(Δx_1)}{dt}+kΔx_1= -AΔP_{ext}(t)-(F_{stat}-F_{kin})$$

I'm having trouble seeing the significance of the change of including ΔP_ext(t), since from what I see the homogeneous equation would stay the same as if ΔP_ext(t) = 0, and only steady state solution changes which will only affect the "shift" of the transient stage/initial compression.

Thank you very much

 

[Chestermiller]

Were you referring to the following relationship for ∂h/∂P:

$$\frac{∂h}{∂P} = [V-T(\frac{∂V}{∂T})_P]dP$$

Does the above only applied to fluids (the book appear to imply this) or any phased material?

This applies to solids, liquids, and gases.

Realistically, what is the "path" that is acceptable to perform the integral? I assume that we cannot just integrate from 1 to 0 atm at constant 298K and then from 298K to 0K at constant 0 atm or vice versa? Typically at 1 atm, how low must the temperature go before we begin to deviate from ideal gas?

You can integrate over whatever path you wish. At 1 atm and below, the behavior of most gases is ideal, and for liquids and solids, the pressure term is usually small.

I don't understand your obsession with trying to integrate down to absolute zero. You seem to be spending a lot of your valuable thinking time on something that, to me, doesn't seem very important practically.

For quasistatic compression, many textbooks describes it generally as increasing external pressure by dP -> piston compresses -> wait until equilibrium is established -> repeat.

This seems to be referring to scenarios in which the external pressure is applied in small discrete increments. If you work this out using our equations, you will find that, as you consider cases in which the increments become smaller and smaller, the amount of irreversibility from the damping combined with the inertia of the piston becomes less and less and approaches the exact quasistatic case of no net force.

By this logic I feel that the piston must always be accelerating since there is always a net external dP acting on the piston, is this correct?

Try doing the analysis using small incremental increases in the external pressure followed by re-equilibration in-between, and see for yourself what you get. (To make things simple, do it for the case of no friction). You have a math model to work with. See what it tells you.

I had a feeling after further thought that a constant velocity quasistatic compression is not achievable since there is always a need for continuous net force?

Just plug in a constant velocity into the differential equation, and see what it tells you. You will find that you can maintain a constant velocity without having a net force. Understand that Pext is not the net force.

With regards to the original equation copied below:

$$m\frac{d^2(Δx_1)}{dt^2}+C\frac{d(Δx_1)}{dt}+kΔx_1= -AΔP_{ext}(t)-(F_{stat}-F_{kin})$$

I'm having trouble seeing the significance of the change of including ΔP_ext(t), since from what I see the homogeneous equation would stay the same as if ΔP_ext(t) = 0, and only steady state solution changes which will only affect the "shift" of the transient stage/initial compression.

Sorry, I don't understand this question.

Chet

 

[Red_CCF]

You can integrate over whatever path you wish. At 1 atm and below, the behavior of most gases is ideal, and for liquids and solids, the pressure term is usually small.

I don't understand your obsession with trying to integrate down to absolute zero. You seem to be spending a lot of your valuable thinking time on something that, to me, doesn't seem very important practically.

I just wanted to get an theoretical understanding of how properties are related between standard reference state and another arbitrary reference state (absolute zero just came to mind first as an example). When you say that I can integrate along any path, does that mean the path I proposed is valid, since it goes through some states where I don't think is physically possible (i.e. no pressure at finite temperature)?

This seems to be referring to scenarios in which the external pressure is applied in small discrete increments. If you work this out using our equations, you will find that, as you consider cases in which the increments become smaller and smaller, the amount of irreversibility from the damping combined with the inertia of the piston becomes less and less and approaches the exact quasistatic case of no net force.

Try doing the analysis using small incremental increases in the external pressure followed by re-equilibration in-between, and see for yourself what you get. (To make things simple, do it for the case of no friction). You have a math model to work with. See what it tells you.

How come the quasistatic case is said to have no net force?

For the case of no friction, at the first time increment when ΔP_ext(0) = 0 and ΔP_ext(dt) = δP I have the following from reducing the equation:

$$m\frac{d^2(Δx_1)}{dt^2}+C\frac{d(Δx_1)}{dt}+kΔx_1= -AΔP_ext(t) = -AδP$$

I am a little lost after this stage, as I am unsure how to add the re-equilibration stage into the equation and that as δP -> dP that ma + cv -> 0.

Just plug in a constant velocity into the differential equation, and see what it tells you. You will find that you can maintain a constant velocity without having a net force. Understand that Pext is not the net force.

Chet

The net force I was referring to was the difference between P_ext and the internal gas pressure. From previous equations at the first time increment, I concluded that for constant velocity, AdP = -kΔx₁ - Cv_piston so v = -1/C(AdP + kΔx₁). I believe that Δx₁ < 0 as t > 0, and for v to stay constant this implies that the difference P_ext and Δx is constant (plotted over time the two have the same trend). This seems to make sense intuitively since kΔx is analogous to gas pressure which for quasistatic process should balance the applied pressure.

Thank you very much

 

[Chestermiller]

I just wanted to get an theoretical understanding of how properties are related between standard reference state and another arbitrary reference state (absolute zero just came to mind first as an example). When you say that I can integrate along any path, does that mean the path I proposed is valid, since it goes through some states where I don't think is physically possible (i.e. no pressure at finite temperature)?



How come the quasistatic case is said to have no net force?

For the case of no friction, at the first time increment when ΔP_ext(0) = 0 and ΔP_ext(dt) = δP I have the following from reducing the equation:

$$m\frac{d^2(Δx_1)}{dt^2}+C\frac{d(Δx_1)}{dt}+kΔx_1= -AΔP_ext(t) = -AδP$$

I am a little lost after this stage, as I am unsure how to add the re-equilibration stage into the equation and that as δP -> dP that ma + cv -> 0.



The net force I was referring to was the difference between P_ext and the internal gas pressure. From previous equations at the first time increment, I concluded that for constant velocity, AdP = -kΔx₁ - Cv_piston so v = -1/C(AdP + kΔx₁). I believe that Δx₁ < 0 as t > 0, and for v to stay constant this implies that the difference P_ext and Δx is constant (plotted over time the two have the same trend). This seems to make sense intuitively since kΔx is analogous to gas pressure which for quasistatic process should balance the applied pressure.

Thank you very much

I can help you figure out the answers to all these questions, but it will be in a couple of days. Grandchildren visiting!

Chet

 

[Chestermiller]

I just wanted to get an theoretical understanding of how properties are related between standard reference state and another arbitrary reference state (absolute zero just came to mind first as an example). When you say that I can integrate along any path, does that mean the path I proposed is valid, since it goes through some states where I don't think is physically possible (i.e. no pressure at finite temperature)?

How come the quasistatic case is said to have no net force?

For the case of no friction, at the first time increment when ΔP_ext(0) = 0 and ΔP_ext(dt) = δP I have the following from reducing the equation:

$$m\frac{d^2(Δx_1)}{dt^2}+C\frac{d(Δx_1)}{dt}+kΔx_1= -AΔP_ext(t) = -AδP$$

I am a little lost after this stage, as I am unsure how to add the re-equilibration stage into the equation and that as δP -> dP that ma + cv -> 0.

The net force I was referring to was the difference between P_ext and the internal gas pressure. From previous equations at the first time increment, I concluded that for constant velocity, AdP = -kΔx₁ - Cv_piston so v = -1/C(AdP + kΔx₁). I believe that Δx₁ < 0 as t > 0, and for v to stay constant this implies that the difference P_ext and Δx is constant (plotted over time the two have the same trend). This seems to make sense intuitively since kΔx is analogous to gas pressure which for quasistatic process should balance the applied pressure.

Thank you very much

We are going to answer your questions by doing a little modeling. There are three basic principles that are important in doing modeling:

1. Start simple
2. Start simple
3. Start simple

Why is it so important to start simple? Because, if you can't solve the simpler versions of your problem, you certainly won't be able to solve the more complicated versions. And, after you solve a simpler version of a problem, you will already have some results under your belt to compare against.

I'm going to formulate two problems for you to work on, representing simpler versions of what you are asking.

Problem A: In your thermo tables, they give two different values for the heat of formation of water at 25 C and 1 atm. One of these is for liquid water, and the other is for the hypothetical state of water vapor. How do they get the value for the hypothetical state of water vapor from the value for liquid water? That is, how do they get the change in enthalpy from liquid water at 1 and 25C to the hypothetical state of water vapor at 1 atm and 25C?

Problem B: (part 1)
I'm going to reformulate the equation relating Pext(t) to x(t) (no friction and no piston mass) in a slightly different form:
$$AP_{ext}(t)=AP_{ext}(0)-C\frac{dx}{dt}-kx$$
where now, x is the displacement relative to the length of the spring at equilibrium at P_ext(0), and AP_ext(0)=P_Ei is the imposed external force at time zero. At time t = 0, the imposed external load is suddenly changed to P_Ef, and held at that value for all subsequent time. Please solve for x(t) as a function of P_Ei, P_Ef, t, k, and C.

After you solve for this, we will look at the amount of work done by the external force, and the contributions of the spring and the damper to that work. We will then subdivide the load change P_Ef-P_Ei into smaller incremental steps, with equilibration between the sequential steps to see how the total amount of work done changes as the number of steps increases. We will compare this with the quasistatic result.

Chet

 

[Chestermiller]

It looks like Red_CCF is no longer participating in this thread. Is there anyone else out there who has been following this thread, and who would like me to complete the solution to Problem B? If not, I'll end it here.

Chet

 

[Red_CCF]

It looks like Red_CCF is no longer participating in this thread. Is there anyone else out there who has been following this thread, and who would like me to complete the solution to Problem B? If not, I'll end it here.

Chet

Hi Chet

My apologies I got a little caught up with work the last week.

We are going to answer your questions by doing a little modeling. There are three basic principles that are important in doing modeling:

1. Start simple
2. Start simple
3. Start simple

Why is it so important to start simple? Because, if you can't solve the simpler versions of your problem, you certainly won't be able to solve the more complicated versions. And, after you solve a simpler version of a problem, you will already have some results under your belt to compare against.

I'm going to formulate two problems for you to work on, representing simpler versions of what you are asking.

Problem A: In your thermo tables, they give two different values for the heat of formation of water at 25 C and 1 atm. One of these is for liquid water, and the other is for the hypothetical state of water vapor. How do they get the value for the hypothetical state of water vapor from the value for liquid water? That is, how do they get the change in enthalpy from liquid water at 1 and 25C to the hypothetical state of water vapor at 1 atm and 25C?

Is there a reason that water vapour at 25C and 1atm is considered hypothetical (since it actually exists)?

My impression was that water vapour heat of formation is equal to the heat of formation of liquid water plus the heat of vaporization of liquid water at 25C and 1atm (h_f,vapour = h_f,liquid+h_fg).

Problem B: (part 1)
I'm going to reformulate the equation relating Pext(t) to x(t) (no friction and no piston mass) in a slightly different form:
$$AP_{ext}(t)=AP_{ext}(0)-C\frac{dx}{dt}-kx$$
where now, x is the displacement relative to the length of the spring at equilibrium at P_ext(0), and AP_ext(0)=P_Ei is the imposed external force at time zero. At time t = 0, the imposed external load is suddenly changed to P_Ef, and held at that value for all subsequent time. Please solve for x(t) as a function of P_Ei, P_Ef, t, k, and C.

After you solve for this, we will look at the amount of work done by the external force, and the contributions of the spring and the damper to that work. We will then subdivide the load change P_Ef-P_Ei into smaller incremental steps, with equilibration between the sequential steps to see how the total amount of work done changes as the number of steps increases. We will compare this with the quasistatic result.

Chet

If I am interpreting Pext(t) correct, it is equal to P_Ei at t = 0 and equal to P_Ef for t > 0? If so and taking x(0) = 0:

$$x(t) = \frac{(P_{Ei}-P_{Ef})}{k}(1-\frac{k}{e^{\frac{kt}{c}}})$$

Thank you very much

 

[Chestermiller]

Hi Chet
Is there a reason that water vapour at 25C and 1atm is considered hypothetical (since it actually exists)?

Wow, I'm glad we're started simple. Pure water vapor at 1 atm pressure is not a stable equilibrium state for water at 25C. At 25C, only pressures less than the equilibrium vapor pressure are stable for water vapor. For liquid water at 25C, only pressures higher than the equilibrium vapor pressure are stable.

To get the change in enthalpy in going from liquid water at 25C and 1 atm. to the hypothetical state of water vapor at 25C and 1 atm, you first calculate the enthalpy change for liquid water in going from 25C and 1 atm. to 25C and the equilibrium vapor pressure. Then you add the heat of vaporization at 25 C and the equilibrium vapor pressure. Then, you recognize that, for an ideal gas, the enthalpy is independent of pressure. So there is no change in enthalpy in going from the equilibrium vapor pressure to a the hypothetical state at 1 atm. Please run this calculation and check to confirm that, if you start with the enthalpy value for liquid water at 25 C and 1 atm and follow this procedure, you end up with the value for water vapor in the hypothetical state of water vapor at 25C and 1 atm that is listed in your table.

If I am interpreting Pext(t) correct, it is equal to P_Ei at t = 0 and equal to P_Ef for t > 0? If so and taking x(0) = 0:

$$x(t) = \frac{(P_{Ei}-P_{Ef})}{k}(1-\frac{k}{e^{\frac{kt}{c}}})$$

This is not what I get. I obtain:
$$x(t) = \frac{(P_{Ei}-P_{Ef})}{k}(1-e^{-\frac{kt}{c}})$$
For part 2 in problem B, I would like you to start with the force balance on the piston (presented in Problem B statement), and multiply both sides of the equation by dx/dt. Then I would like you to integrate each term in the resulting equation from t = 0 to t = ∞. Please do this for both the left side of the equation as well as for each of the terms individually on the right side of the equation. The integration of the left side of the equation will tell us the total amount of work that is done on the system. The integration of the terms on the right side will tell us the individual contributions of the elastic spring (analogous to the PV behavior of the gas) and the damper (analogous to the viscous dissipation in the gas) to the total amount of work.

Chet

 

[Red_CCF]

Wow, I'm glad we're started simple. Pure water vapor at 1 atm pressure is not a stable equilibrium state for water at 25C. At 25C, only pressures less than the equilibrium vapor pressure are stable for water vapor. For liquid water at 25C, only pressures higher than the equilibrium vapor pressure are stable.

My bad I was thinking about a water-air system (not pure water system) for some reason. From the steam table the equilibrium vapor pressure you are referring to is 3.17kPa?

To get the change in enthalpy in going from liquid water at 25C and 1 atm. to the hypothetical state of water vapor at 25C and 1 atm, you first calculate the enthalpy change for liquid water in going from 25C and 1 atm. to 25C and the equilibrium vapor pressure. Then you add the heat of vaporization at 25 C and the equilibrium vapor pressure. Then, you recognize that, for an ideal gas, the enthalpy is independent of pressure. So there is no change in enthalpy in going from the equilibrium vapor pressure to a the hypothetical state at 1 atm. Please run this calculation and check to confirm that, if you start with the enthalpy value for liquid water at 25 C and 1 atm and follow this procedure, you end up with the value for water vapor in the hypothetical state of water vapor at 25C and 1 atm that is listed in your table.

I ran into a little problem with this one.

Since the table in my combustion book is referenced at standard state the sensible component is zero and I get h(25C, 1atm) = h_f = -241845kJ/kmol.

I was taught to approximate liquid water enthalpy at a specific temperature to the enthalpy of saturated liquid at the same temperature. In this case the enthapies of the liquid is pressure independent and the change in enthalpies from 25C 1atm to 25C equilibrium pressure is zero; I'm not sure if there's a better approach than this. The only thing left is the h_fg = 2409.8kJ/kg = 43412.5kJ/kmol. However, the reference state for this value is that for liquid water at 0 C, 1atm and I'm not sure how I would convert it to standard state reference.

This is not what I get. I obtain:
$$x(t) = \frac{(P_{Ei}-P_{Ef})}{k}(1-e^{-\frac{kt}{c}})$$
For part 2 in problem B, I would like you to start with the force balance on the piston (presented in Problem B statement), and multiply both sides of the equation by dx/dt. Then I would like you to integrate each term in the resulting equation from t = 0 to t = ∞. Please do this for both the left side of the equation as well as for each of the terms individually on the right side of the equation. The integration of the left side of the equation will tell us the total amount of work that is done on the system. The integration of the terms on the right side will tell us the individual contributions of the elastic spring (analogous to the PV behavior of the gas) and the damper (analogous to the viscous dissipation in the gas) to the total amount of work.

Chet

Sorry the k was a typo and should be a 1.

After mutiplying both sides by dx/dt I get:

$$P_{Ef}\frac{(P_{Ei}-P_{Ef})}{C}e^{\frac{-kt}{C}}=P_{Ei}\frac{(P_{Ei}-P_{Ef})}{C}e^{\frac{-kt}{C}}-\frac{(P_{Ei}-P_{Ef})^2}{C}e^{\frac{-2kt}{C}}-\frac{(P_{Ei}-P_{Ef})^2}{C}(e^{\frac{-kt}{C}}-e^{\frac{-2kt}{C}})$$
Integrating from t = 0 to t = infinity, without simplification

$$P_{Ef}\frac{(P_{Ei}-P_{Ef})}{k}=P_{Ei}\frac{(P_{Ei}-P_{Ef})}{k}-\frac{(P_{Ei}-P_{Ef})^2}{2k}-\frac{(P_{Ei}-P_{Ef})^2}{2k}$$
From the above it looks like the damper and spring does equal amount of work. What is the significance of the sign difference between the damper/spring and the initial/final pressures?

Thank you very much

 

[Chestermiller]

My bad I was thinking about a water-air system (not pure water system) for some reason. From the steam table the equilibrium vapor pressure you are referring to is 3.17kPa?

Yes.

I ran into a little problem with this one.

Since the table in my combustion book is referenced at standard state the sensible component is zero and I get h(25C, 1atm) = h_f = -241845kJ/kmol.

I was taught to approximate liquid water enthalpy at a specific temperature to the enthalpy of saturated liquid at the same temperature. In this case the enthapies of the liquid is pressure independent and the change in enthalpies from 25C 1atm to 25C equilibrium pressure is zero; I'm not sure if there's a better approach than this. The only thing left is the h_fg = 2409.8kJ/kg = 43412.5kJ/kmol. However, the reference state for this value is that for liquid water at 0 C, 1atm and I'm not sure how I would convert it to standard state reference.

The heats for formation table on the internet gives:

Water vapor in hypothetical state of 25 C and 1 atm: -241.818 kJ/mol

Liquid water at 25C and 1 atm: -285.8 kJ/mol

The heat of vaporization of water at 25C is 43.99 kJ/mol

If one also wishes to include the change in enthalpy in dropping the pressure of liquid water from 1 atm to the equilibrium vapor pressure at 25C, one would use:
$$ΔH=V(1-αT)ΔP$$
where V is the specific volume of liquid water at 25C, α is the coefficient of volumetric thermal expansion of liquid water at 25C, and ΔP is the pressure difference between the equilibrium vapor pressure and 1 atm. See what magnitude this term contributes, compared to the heat of vaporization.

Sorry the k was a typo and should be a 1.

After mutiplying both sides by dx/dt I get:

$$P_{Ef}\frac{(P_{Ei}-P_{Ef})}{C}e^{\frac{-kt}{C}}=P_{Ei}\frac{(P_{Ei}-P_{Ef})}{C}e^{\frac{-kt}{C}}-\frac{(P_{Ei}-P_{Ef})^2}{C}e^{\frac{-2kt}{C}}-\frac{(P_{Ei}-P_{Ef})^2}{C}(e^{\frac{-kt}{C}}-e^{\frac{-2kt}{C}})$$
Integrating from t = 0 to t = infinity, without simplification

$$P_{Ef}\frac{(P_{Ei}-P_{Ef})}{k}=P_{Ei}\frac{(P_{Ei}-P_{Ef})}{k}-\frac{(P_{Ei}-P_{Ef})^2}{2k}-\frac{(P_{Ei}-P_{Ef})^2}{2k}$$
From the above it looks like the damper and spring does equal amount of work. What is the significance of the sign difference between the damper/spring and the initial/final pressures?

In the first term on the right hand side, if we write
$$P_{Ei}=\frac{P_{Ei}+P_{Ef}}{2}+\frac{P_{Ei}-P_{Ef}}{2}$$
We obtain:
$$P_{Ef}\frac{(P_{Ei}-P_{Ef})}{k}=\frac{P_{Ei}^2-P_{Ef}^2}{2k}-\frac{(P_{Ei}-P_{Ef})^2}{2k}$$
From this, it follows that:
Total work done by surroundings to compress spring and damper = $P_{Ef}\frac{(P_{Ef}-P_{Ei})}{k}$
Work done by surroundings to compress spring = $\frac{P_{Ef}^2-P_{Ei}^2}{2k}$
Work done by surroundings to compress damper = $\frac{(P_{Ef}-P_{Ei})^2}{2k}$

Note that the work done by the surroundings to compress the spring is just the elastic energy stored in the spring, and this is analogous to the reversible P-V work required to compress the gas. The work done by the surroundings to compress the damper is positive definite, and this is analogous to the viscous energy dissipation.

Next, we are going to see how these results are affected if we do the compression in a two-step process, rather than the single large step just completed.

Step 1: Compress the gas from pressure P_Ei to pressure (P_Ei+P_Ef)/2, allowing the system to re-equilibrate at the new pressure
Step 2: Compress the gas from pressure (P_Ei+P_Ef)/2 to pressure P_Ef, allowing the system to re-equilibrate at the new pressure
I want you to determine the following for each of these Steps:
(a) The total work done by the surroundings to compress the spring and damper
(b) The work done by the surroundings to compress the spring
(c) The work done by the surroundings to compress the damper

(You don't need to do the integrations again to obtain these results. You can use the algebraic results from the final equation describing the work terms in the single step process).

Then, I'd like you to determine the sum of these quantities over the combination of the two steps. We can then compare the results in going from P_Ei to P_Ef in a two step process with going directly from P_Ei to P_Ef in a single step process. The results, particularly with respect to the work to compress the damper will be very revealing.

Chet

 

[Red_CCF]

The heats for formation table on the internet gives:

Water vapor in hypothetical state of 25 C and 1 atm: -241.818 kJ/mol

Liquid water at 25C and 1 atm: -285.8 kJ/mol

The heat of vaporization of water at 25C is 43.99 kJ/mol

If one also wishes to include the change in enthalpy in dropping the pressure of liquid water from 1 atm to the equilibrium vapor pressure at 25C, one would use:
$$ΔH=V(1-αT)ΔP$$
where V is the specific volume of liquid water at 25C, α is the coefficient of volumetric thermal expansion of liquid water at 25C, and ΔP is the pressure difference between the equilibrium vapor pressure and 1 atm. See what magnitude this term contributes, compared to the heat of vaporization.

From the above equation (using alpha at 20oC):

$$ΔH=V(1-αT)ΔP = 1.0029*10^{-3}(1-207x10^{-6}*25)*(101.3-3.17)= 0.0979kJ/kg =0.00176kJ/mol$$
I wasn't sure how to deal with the αT term since α is given per degree of temperature change (so 1K = 1C) while I used the absolute temperature, although numerically it didn't make much difference. Also assuming that specific volume change with pressure is small and using the saturated liquid (25C) specific volume. In any case it looks like the liquid enthalpy change is insignificant.

From where was the enthalpy of liquid water at 25C and 1atm (-285.8 kJ/mol) found? My textbook and the internet gave me very different values (and all positive).

In the first term on the right hand side, if we write
$$P_{Ei}=\frac{P_{Ei}+P_{Ef}}{2}+\frac{P_{Ei}-P_{Ef}}{2}$$
We obtain:
$$P_{Ef}\frac{(P_{Ei}-P_{Ef})}{k}=\frac{P_{Ei}^2-P_{Ef}^2}{2k}-\frac{(P_{Ei}-P_{Ef})^2}{2k}$$
From this, it follows that:
Total work done by surroundings to compress spring and damper = $P_{Ef}\frac{(P_{Ef}-P_{Ei})}{k}$
Work done by surroundings to compress spring = $\frac{P_{Ef}^2-P_{Ei}^2}{2k}$
Work done by surroundings to compress damper = $\frac{(P_{Ef}-P_{Ei})^2}{2k}$

Note that the work done by the surroundings to compress the spring is just the elastic energy stored in the spring, and this is analogous to the reversible P-V work required to compress the gas. The work done by the surroundings to compress the damper is positive definite, and this is analogous to the viscous energy dissipation.

I noticed that the integrated equation was re-arranged into this form based on the description for components of work done by surroundings:
$$-P_{Ef}\frac{(P_{Ef}-P_{Ei})}{k}=-\frac{P_{Ef}^2-P_{Ei}^2}{2k}-\frac{(P_{Ef}-P_{Ei})^2}{2k}$$
Is there a reason to extract the minus in front of every equation and have P_Ef as the leading term?

Next, we are going to see how these results are affected if we do the compression in a two-step process, rather than the single large step just completed.

Step 1: Compress the gas from pressure P_Ei to pressure (P_Ei+P_Ef)/2, allowing the system to re-equilibrate at the new pressure
Step 2: Compress the gas from pressure (P_Ei+P_Ef)/2 to pressure P_Ef, allowing the system to re-equilibrate at the new pressure
I want you to determine the following for each of these Steps:
(a) The total work done by the surroundings to compress the spring and damper
(b) The work done by the surroundings to compress the spring
(c) The work done by the surroundings to compress the damper

(You don't need to do the integrations again to obtain these results. You can use the algebraic results from the final equation describing the work terms in the single step process).

Then, I'd like you to determine the sum of these quantities over the combination of the two steps. We can then compare the results in going from P_Ei to P_Ef in a two step process with going directly from P_Ei to P_Ef in a single step process. The results, particularly with respect to the work to compress the damper will be very revealing.

Chet

For Step 1:
a)$\frac{P_{Ef}^2-P_{Ei}^2}{4k}$
b)$\frac{(P_{Ef}+P_{Ei})^2-4P_{Ei}^2}{8k}$
c)$\frac{(P_{Ef}-P_{Ei})^2}{8k}$

For Step 2:
a)$\frac{ P_{Ef}(P_{Ef}-P_{Ei})}{2k}$
b)$\frac{4P_{Ef}^2-(P_{Ef}+P_{Ei})^2}{8k}$
c)$\frac{(P_{Ef}-P_{Ei})^2}{8k}$

The Sum:
a)$\frac{3P_{Ef}^2-P_{Ei}^2-2P_{Ef}P_{Ei}}{4k}$
b)$\frac{P_{Ef}^2-P_{Ei}^2}{2k}$
c)$\frac{(P_{Ef}-P_{Ei})^2}{4k}$

It looks like that the work performed by the damper decreased by a factor of two (no. of steps) while that by the spring stayed constant.


Thank you very much

 

[Chestermiller]

From the above equation (using alpha at 20oC):

$$ΔH=V(1-αT)ΔP = 1.0029*10^{-3}(1-207x10^{-6}*25)*(101.3-3.17)= 0.0979kJ/kg =0.00176kJ/mol$$
I wasn't sure how to deal with the αT term since α is given per degree of temperature change (so 1K = 1C) while I used the absolute temperature, although numerically it didn't make much difference.

You should be using the absolute temperature in the calculation.

From where was the enthalpy of liquid water at 25C and 1atm (-285.8 kJ/mol) found? My textbook and the internet gave me very different values (and all positive).

This is the heat of formation of liquid water at 25C. I found it in an internet table of heats of formation. It is also, of course, the heat of combustion of H2 and O2 at 25C and 1 atm. The value is consistent with the heat of formation of water vapor at the hypothetical state of 25C and 1 atm. together with the heat of vaporization of water at 25C. The other tables you have seen (like steam tables) were probably using a different reference state, like liquid water at 0C.

I noticed that the integrated equation was re-arranged into this form based on the description for components of work done by surroundings:
$$-P_{Ef}\frac{(P_{Ef}-P_{Ei})}{k}=-\frac{P_{Ef}^2-P_{Ei}^2}{2k}-\frac{(P_{Ef}-P_{Ei})^2}{2k}$$
Is there a reason to extract the minus in front of every equation and have P_Ef as the leading term?

Not really. Just personal bias, I guess. For compression, I like to think about the work done to compress the system, rather than the negative work done by the system on the surroundings.

For Step 1:
a)$\frac{P_{Ef}^2-P_{Ei}^2}{4k}$
b)$\frac{(P_{Ef}+P_{Ei})^2-4P_{Ei}^2}{8k}$
c)$\frac{(P_{Ef}-P_{Ei})^2}{8k}$

For Step 2:
a)$\frac{ P_{Ef}(P_{Ef}-P_{Ei})}{2k}$
b)$\frac{4P_{Ef}^2-(P_{Ef}+P_{Ei})^2}{8k}$
c)$\frac{(P_{Ef}-P_{Ei})^2}{8k}$

The Sum:
a)$\frac{3P_{Ef}^2-P_{Ei}^2-2P_{Ef}P_{Ei}}{4k}$
b)$\frac{P_{Ef}^2-P_{Ei}^2}{2k}$
c)$\frac{(P_{Ef}-P_{Ei})^2}{4k}$

It looks like that the work performed by the damper decreased by a factor of two (no. of steps) while that by the spring stayed constant.

Yes. The work performed on the spring is what we identify as the quasistatic work. Note that, as you indicated, when we went from 1 to 2 discrete steps, the quasistatic work remained the same, while the dissipative work decreased, and the total amount of work decreased. What do you think would happen if we went to a very large number of steps, holding the initial and final pressures of the sequence constant? How would the total amount of work compare with the quasistatic work as the number of steps became infinite. What would happen to the total dissipative work?

Chet

 

[Red_CCF]

You should be using the absolute temperature in the calculation.


This is the heat of formation of liquid water at 25C. I found it in an internet table of heats of formation. It is also, of course, the heat of combustion of H2 and O2 at 25C and 1 atm. The value is consistent with the heat of formation of water vapor at the hypothetical state of 25C and 1 atm. together with the heat of vaporization of water at 25C. The other tables you have seen (like steam tables) were probably using a different reference state, like liquid water at 0C.

From my steam tables the enthalpy of saturated liquid at 25C is 104.9kJ/kg = 1.89kJ/mol. Is this value equivalent to the heat of formation of liquid water -285.8kJ/mol relative to standard state (i.e. they have the same "absolute" enthalpy) if pressure effect on liquid enthalpy is neglected? What would we need to theoretically convert from one reference state to the other?

Yes. The work performed on the spring is what we identify as the quasistatic work. Note that, as you indicated, when we went from 1 to 2 discrete steps, the quasistatic work remained the same, while the dissipative work decreased, and the total amount of work decreased. What do you think would happen if we went to a very large number of steps, holding the initial and final pressures of the sequence constant? How would the total amount of work compare with the quasistatic work as the number of steps became infinite. What would happen to the total dissipative work?

Chet

I would imagine that the quasistatic/spring work remain constant, the dissipative work decrease by a factor of 2ⁿ (for n steps) and the total dissipative work approaches the quasistatic/spring work.

Thank you very much

 

[Chestermiller]

From my steam tables the enthalpy of saturated liquid at 25C is 104.9kJ/kg = 1.89kJ/mol. Is this value equivalent to the heat of formation of liquid water -285.8kJ/mol relative to standard state (i.e. they have the same "absolute" enthalpy) if pressure effect on liquid enthalpy is neglected?

25 C is the standard state for liquid water. If we are talking about "absolute enthalpies," -285.8kJ/mol is the absolute enthalpy of liquid water at 25 C minus the absolute enthalpies of H2 and O2 at 25C. The 1.89 kJ/mol is the absolute enthalpy of liquid water at 25 C minus the absolute enthalpy of liquid water at 0C.

What would we need to theoretically convert from one reference state to the other?

You will notice that one of these definitions includes the absolute enthalpies of H2 and O2, while the other does not. So to convert from one to the other, our equation for converting would have to involve the absolute enthalpies of H2 and O2.

I would imagine that the quasistatic/spring work remain constant, the dissipative work decrease by a factor of 2ⁿ (for n steps) and the total dissipative work approaches the quasistatic/spring work.

I think you meant to say that the total work approaches the quasistaic/spring work.

You said, "I would imagine...". If you are not sure, please redo the analysis by going to 3 equal discrete steps, instead of two. If you are still not sure, please redo the analysis by by going to 4 equal discrete steps. I want you to be sure.

Once you confirm that you're sure, I will show you how this plays out when we do the exact same analysis for an actual ideal gas being compressed (including determination of the amount of work done to overcome viscous dissipation and irreversible heat conduction).

Chet

 

[Red_CCF]

25 C is the standard state for liquid water. If we are talking about "absolute enthalpies," -285.8kJ/mol is the absolute enthalpy of liquid water at 25 C minus the absolute enthalpies of H2 and O2 at 25C. The 1.89 kJ/mol is the absolute enthalpy of liquid water at 25 C minus the absolute enthalpy of liquid water at 0C.

You will notice that one of these definitions includes the absolute enthalpies of H2 and O2, while the other does not. So to convert from one to the other, our equation for converting would have to involve the absolute enthalpies of H2 and O2.

So both -285.5kJ/mol and 1.89kJ/mol are calculated by subtracting the same "absolute enthalpy" of H2O (at 25C) with different reference state enthalpies (liquid water at 0C and H2/O2 absolute enthalpies at 25C)? Does this mean that to convert the enthalpy from one reference state to the other we just add/subtract 287.4kJ/mol (at 25C)?

I think you meant to say that the total work approaches the quasistaic/spring work.

You said, "I would imagine...". If you are not sure, please redo the analysis by going to 3 equal discrete steps, instead of two. If you are still not sure, please redo the analysis by by going to 4 equal discrete steps. I want you to be sure.

Once you confirm that you're sure, I will show you how this plays out when we do the exact same analysis for an actual ideal gas being compressed (including determination of the amount of work done to overcome viscous dissipation and irreversible heat conduction).

Chet

Yes I meant the total work. I re-did the calculation for 3 equal steps P_Ei to (P_Ef + 2P_Ei)/3 to (2P_Ef + P_Ei)/3 to P_Ef as below:

a)$\frac{3P_{Ef}^2-3P_{Ei}^2+(P_{Ef}^2-P_{Ei}^2)}{6k}$
b)$\frac{P_{Ef}^2-P_{Ei}^2}{2k}$
c)$\frac{(P_{Ef}-P_{Ei})^2}{6k}$

and the results are as expected.

Thank you very much

 

[Chestermiller]

So both -285.5kJ/mol and 1.89kJ/mol are calculated by subtracting the same "absolute enthalpy" of H2O (at 25C) with different reference state enthalpies (liquid water at 0C and H2/O2 absolute enthalpies at 25C)? Does this mean that to convert the enthalpy from one reference state to the other we just add/subtract 287.4kJ/mol (at 25C)?

Sure.

Yes I meant the total work. I re-did the calculation for 3 equal steps P_Ei to (P_Ef + 2P_Ei)/3 to (2P_Ef + P_Ei)/3 to P_Ef as below:

a)$\frac{3P_{Ef}^2-3P_{Ei}^2+(P_{Ef}^2-P_{Ei}^2)}{6k}$
b)$\frac{P_{Ef}^2-P_{Ei}^2}{2k}$
c)$\frac{(P_{Ef}-P_{Ei})^2}{6k}$

and the results are as expected.

Good job. I hope you can now see how modelling can help us improve our understanding and remove uncertainties.

Now, let's look at the case of an ideal gas.
Problem Statement: We have 1 mole of an ideal gas in a cylinder at temperature T. The cylinder is in contact with a constant temperature reservoir at temperature T. (This means that the temperature at the interface between the system and the surroundings is always controlled to be constant at T). There is a massless piston, and, prior to time t = 0, the force we are applying to the piston is AP_Ei. At time zero, we suddenly raise the force on the piston to AP_Ef and control it so that it is constant at this value for all subsequent times t > 0. We allow the system to re-equilibrate at the new pressure. During the compressional deformation that occurs, the pressure and temperature within the gas may become non-uniform, but, in the end, the equilibrated gas pressure will be P_Ef and the equilibrated gas temperature will be T.

Part 1.
What is the work done by the piston on the gas? What would the quasistatic work have been if the compression had been carried out reversibly? Algebraically, what is the difference between these two amounts of work?

Chet

 

[Red_CCF]

Sure.

Good job. I hope you can now see how modelling can help us improve our understanding and remove uncertainties.

Now, let's look at the case of an ideal gas.
Problem Statement: We have 1 mole of an ideal gas in a cylinder at temperature T. The cylinder is in contact with a constant temperature reservoir at temperature T. (This means that the temperature at the interface between the system and the surroundings is always controlled to be constant at T). There is a massless piston, and, prior to time t = 0, the force we are applying to the piston is AP_Ei. At time zero, we suddenly raise the force on the piston to AP_Ef and control it so that it is constant at this value for all subsequent times t > 0. We allow the system to re-equilibrate at the new pressure. During the compressional deformation that occurs, the pressure and temperature within the gas may become non-uniform, but, in the end, the equilibrated gas pressure will be P_Ef and the equilibrated gas temperature will be T.

Part 1.
What is the work done by the piston on the gas? What would the quasistatic work have been if the compression had been carried out reversibly? Algebraically, what is the difference between these two amounts of work?

Chet

I had some trouble figuring this out. Do I integrate ∫AP_Efdx with the quasistatic case being C = 0? After substituting dx = (P_Ei-P_Ef)/C*e^kt/Cdt, if C = 0 then the solution is undefined? I think the first law (Q = W) should be used somewhere but could not out figure how.

Thanks

 

[Chestermiller]

I had some trouble figuring this out. Do I integrate ∫AP_Efdx with the quasistatic case being C = 0? After substituting dx = (P_Ei-P_Ef)/C*e^kt/Cdt, if C = 0 then the solution is undefined? I think the first law (Q = W) should be used somewhere but could not out figure how.

Thanks

Let's first focus in the left side of the equation , the total work done. The applied force is constant. From the ideal gas law, in terms of the initial and final pressures, what are the initial and final volumes? In terms of the initial and final pressures, what is the total work done?

Chet

 

[Red_CCF]

Let's first focus in the left side of the equation , the total work done. The applied force is constant. From the ideal gas law, in terms of the initial and final pressures, what are the initial and final volumes? In terms of the initial and final pressures, what is the total work done?

Chet

The initial volume would be V_i = nRT/P_Ei and the final volume is V_f = nRT/P_Ef for both cases. So W = PΔV = nRT(1-P_Ef/P_Ei)?

Thanks

 

[Chestermiller]

The initial volume would be V_i = nRT/P_Ei and the final volume is V_f = nRT/P_Ef for both cases. So W = PΔV = nRT(1-P_Ef/P_Ei)?

Thanks

Good (even though I said n is 1 mole). Now, how much would be done if an isothermal reversible path were followed between the same two end points.

Chet

 

[Red_CCF]

Good (even though I said n is 1 mole). Now, how much would be done if an isothermal reversible path were followed between the same two end points.

Chet

For this case, would it be W = ∫pdV = ∫RT/VdV = RTln(V_f/V_i), where V_f/V_i = P_Ei/P_Ef

The difference between the two work terms would be W_non-quasi - W_quasi = RT[1-P_Ef/P_Ei - ln(P_Ei/P_Ef)]. I expect the difference to be positive, but am not sure how to show this in the expression above.

Thank you very much

 

[Chestermiller]

For this case, would it be W = ∫pdV = ∫RT/VdV = RTln(V_f/V_i), where V_f/V_i = P_Ei/P_Ef

The difference between the two work terms would be W_non-quasi - W_quasi = RT[1-P_Ef/P_Ei - ln(P_Ei/P_Ef)]. I expect the difference to be positive, but am not sure how to show this in the expression above.

Thank you very much

Good. So, writing your results in a little different way:
$$-W_{irrev}=RT\left(\frac{P_{Ef}}{P_{Ei}}-1\right)$$
$$-W_{rev}=RT\ln{\left(\frac{P_{Ef}}{P_{Ei}}\right)}$$
So,

$$-(W_{irrev}-W_{rev})=RT\left(\frac{P_{Ef}}{P_{Ei}}-1\right)-RT\ln{\left(\frac{P_{Ef}}{P_{Ei}}\right)}$$

Now, as in the spring/damper problem, the difference between the non-quasistatic (irreversible) work and the quasistatic (reversible) work is equal to the extra amount of compression work required to overcome viscous stresses. So $-(W_{irrev}-W_{rev})$ is the work to overcome viscous stresses.

In our equation for $-W_{irrev}$, we can re-express this equation using the identity:

$$-W_{irrev}=-W_{rev}-(W_{irrev}-W_{rev})=RT\ln{\left(\frac{P_{Ef}}{P_{Ei}}\right)}+RT\left[\left(\frac{P_{Ef}}{P_{Ei}}-1\right)-\ln{\left(\frac{P_{Ef}}{P_{Ei}}\right)}\right]$$

Now,
$$\ln{\left(\frac{P_{Ef}}{P_{Ei}}\right)}=\ln{(P_{Ef})}-\ln{(P_{Ei})}$$
and
$$\frac{P_{Ef}}{P_{Ei}}=e^{(\ln{(P_{Ef})}-\ln{(P_{Ei})})}$$
If we substitute these identities into our equation for $-W_{irrev}$, we obtain;

$$-W_{irrev}=RT(\ln{(P_{Ef})}-\ln{(P_{Ei})}) +RT\left[e^{(\ln{(P_{Ef})}-\ln{(P_{Ei})})}-1 -(\ln{(P_{Ef})}-\ln{(P_{Ei})})\right]$$

The term involving brackets in this equation represents the irreversible contribution of viscous stresses to the total amount of work.

If we expand the term in brackets in a Taylor Series in $(\ln{(P_{Ef})}-\ln{(P_{Ei})})$ and retain terms only up to quadratic terms, we obtain:

$$-W_{irrev}≈RT(\ln{(P_{Ef})}-\ln{(P_{Ei})}) +RT\frac{(\ln{(P_{Ef})}-\ln{(P_{Ei})})^2}{2}$$

The first term on the RHS of this equation is the reversible work, and the second term represents a close approximation to the work required to overcome viscous stresses. Note that both terms are expressed solely in terms of $(\ln{(P_{Ef})}-\ln{(P_{Ei})})$. The sum of the two terms is the irreversible work.

This completes what I wanted to do on the single large-pressure-step problem. Please compare these results with what we obtained for the case where the spring and damper were inside the cylinder, rather than an ideal gas. You will see some very close similarities.

Now, I'd like to go on to a two step process. The problem is exactly the same, except that, in Step 1, we go from P_Ei to $\sqrt{P_{Ei}P_{Ef}}$, and in Step 2, we go from $\sqrt{P_{Ei}P_{Ef}}$ to P_Ef. Determine for each step (a) the irreversible work, (b) the reversible quasistatic work, and (c) the viscous work. Then determine the sums of these over the combination of the two steps. Then compare the results with what we got for the single step process.

Chet

 

[Red_CCF]

Good. So, writing your results in a little different way:
$$-W_{irrev}=RT\left(\frac{P_{Ef}}{P_{Ei}}-1\right)$$
$$-W_{rev}=RT\ln{\left(\frac{P_{Ef}}{P_{Ei}}\right)}$$
So,

$$-(W_{irrev}-W_{rev})=RT\left(\frac{P_{Ef}}{P_{Ei}}-1\right)-RT\ln{\left(\frac{P_{Ef}}{P_{Ei}}\right)}$$

Now, as in the spring/damper problem, the difference between the non-quasistatic (irreversible) work and the quasistatic (reversible) work is equal to the extra amount of compression work required to overcome viscous stresses. So $-(W_{irrev}-W_{rev})$ is the work to overcome viscous stresses.

In our equation for $-W_{irrev}$, we can re-express this equation using the identity:

$$-W_{irrev}=-W_{rev}-(W_{irrev}-W_{rev})=RT\ln{\left(\frac{P_{Ef}}{P_{Ei}}\right)}+RT\left[\left(\frac{P_{Ef}}{P_{Ei}}-1\right)-\ln{\left(\frac{P_{Ef}}{P_{Ei}}\right)}\right]$$

Now,
$$\ln{\left(\frac{P_{Ef}}{P_{Ei}}\right)}=\ln{(P_{Ef})}-\ln{(P_{Ei})}$$
and
$$\frac{P_{Ef}}{P_{Ei}}=e^{(\ln{(P_{Ef})}-\ln{(P_{Ei})})}$$
If we substitute these identities into our equation for $-W_{irrev}$, we obtain;

$$-W_{irrev}=RT(\ln{(P_{Ef})}-\ln{(P_{Ei})}) +RT\left[e^{(\ln{(P_{Ef})}-\ln{(P_{Ei})})}-1 -(\ln{(P_{Ef})}-\ln{(P_{Ei})})\right]$$

The term involving brackets in this equation represents the irreversible contribution of viscous stresses to the total amount of work.

If we expand the term in brackets in a Taylor Series in $(\ln{(P_{Ef})}-\ln{(P_{Ei})})$ and retain terms only up to quadratic terms, we obtain:

$$-W_{irrev}≈RT(\ln{(P_{Ef})}-\ln{(P_{Ei})}) +RT\frac{(\ln{(P_{Ef})}-\ln{(P_{Ei})})^2}{2}$$

The first term on the RHS of this equation is the reversible work, and the second term represents a close approximation to the work required to overcome viscous stresses. Note that both terms are expressed solely in terms of $(\ln{(P_{Ef})}-\ln{(P_{Ei})})$. The sum of the two terms is the irreversible work.

This completes what I wanted to do on the single large-pressure-step problem. Please compare these results with what we obtained for the case where the spring and damper were inside the cylinder, rather than an ideal gas. You will see some very close similarities.

Now, I'd like to go on to a two step process. The problem is exactly the same, except that, in Step 1, we go from P_Ei to $\sqrt{P_{Ei}P_{Ef}}$, and in Step 2, we go from $\sqrt{P_{Ei}P_{Ef}}$ to P_Ef. Determine for each step (a) the irreversible work, (b) the reversible quasistatic work, and (c) the viscous work. Then determine the sums of these over the combination of the two steps. Then compare the results with what we got for the single step process.

Chet

I think each step is identical, so the below would apply to both steps:

a) $-W_{irrev}=RT\left(\frac{\sqrt{P_{Ef}}}{\sqrt{P_{Ei}}}-1\right)$
b) $-W_{rev}=RT\ln{\left(\frac{\sqrt{P_{Ef}}}{\sqrt{P_{Ei}}}\right)}$
c) $-W_{visc}=RT\frac{(\ln{(P_{Ef})}-\ln{(P_{Ei})})^2}{8}$

The entire process should be each of the above multiplied by two

a) $-W_{irrev}=2RT\left(\frac{\sqrt{P_{Ef}}}{\sqrt{P_{Ei}}}-1\right)$
b) $-W_{rev}=RT\ln{\left(\frac{P_{Ef}}{P_{Ei}}\right)}$
c) $-W_{visc}=RT\frac{(\ln{(P_{Ef})}-\ln{(P_{Ei})})^2}{4}$

The reversible work stayed the same while the viscous stress from the irreversible process was halved with the additional step much like in the spring-damper case.

Thank you very much

 

[Chestermiller]

OK. Good job.

So, to summarize, as we increase the number of discrete intervals, the total work approaches the reversible quasistaic work, and the work to overcome viscous stresses becomes smaller and smaller, ultimately approaching zero.

I think this answers your original question.

Chet

 

[Red_CCF]

OK. Good job.

So, to summarize, as we increase the number of discrete intervals, the total work approaches the reversible quasistaic work, and the work to overcome viscous stresses becomes smaller and smaller, ultimately approaching zero.

I think this answers your original question.

Chet

Looking at this process qualitatively, is the quasistatic process essentially: apply a infinitessimal pressure increase -> piston compresses until interface/external pressure balances -> wait for equilibrium -> repeat?

With regards to the piston kinematics (if massless and frictionless), the massless piston means that external/interface must always equal, how does one actually control the piston's motion (moving it initially when increase P_ext by dP and making it stop after some compression so the system can reach equilibrium)?

Thank you very much

 

[Chestermiller]

Looking at this process qualitatively, is the quasistatic process essentially: apply a infinitessimal pressure increase -> piston compresses until interface/external pressure balances -> wait for equilibrium -> repeat?

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?

With regards to the piston kinematics (if massless and frictionless), the massless piston means that external/interface must always equal, how does one actually control the piston's motion (moving it initially when increase P_ext by dP and making it stop after some compression so the system can reach equilibrium)?

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