Question about Flow between Parallel Plates

[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

 

[Red_CCF]

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

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

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

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

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

Chet

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

Thank you very much

 

[Chestermiller]

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

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

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

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

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

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

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

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

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

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

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

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

Chet

 

[Red_CCF]

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

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

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

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

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

Chet

What is oomph?

Thank you very much

 

[Chestermiller]

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

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

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

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

What is oomph?

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

Chet

 

[Red_CCF]

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

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

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

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

Thank you very much

 

[Chestermiller]

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

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

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

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

Chet

 

[Red_CCF]

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

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

Thank you very much

 

[Chestermiller]

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

Thank you very much

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

Chet

 

[Chestermiller]

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

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

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

Chet

 

[Red_CCF]

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

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

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

Chet

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

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

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

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

Thank you very much