Polytropic Piston-cylinder & entropy

[George32]

Homework Statement

We have a piston in a cylinder containing Helium. The initial states are:
P₁=150kPa
T₁= 20°C
V₁=0.5m³

Following a polytropic process, the final states are:
P₂=400kPa
T₂=140°C
V₂ is unknown.

We're also given R=2.0769 kPa.m³.(kg.K)^-1
And C_p=5.1926kJ.(kg.K)^-1
As well as the temperature of the surrounding being 20°C

We are then asked to find:
a) The entropy change inside the system
b) The entropy change in the surroundings
c) Is the process Irreversible, Reversible or Impossible?


A hint is that we need to find n (the polytropic constant)

Homework Equations

PV=RT (1)
$TdS=dh - VdP$ (2)
$dh=CpdT$ (3)
$PV^n = Constant$ (4)

The Attempt at a Solution

I have a solution for part (a) which comes out at -0.2546kJ/K.

I got this after rearranging (1) to give $$\frac{V}{T} = \frac{R}{P}$$ and filling (3) and the new (1) into (2) to give $$ds=\frac{CpdT}{T} - \frac{RdP}{P}$$

Integrating each term between their respective limits to give $$\Delta s = Cp(ln (\frac{T2}{T1})) - R(ln(\frac{P2}{P1}))$$

Now for the process to be polytropic it has to be quasi-static which doesn't necessarily mean reversible but is (as I understand it) usually reversible, so the entropy change of the surroundings would have to be 0.2546 at least. However I don't know how I would calculate this, and the third part makes me question if I should get an answer lower and it will be an impossible process.

Any pointers on how to proceed, as well as perhaps a confirmation of using the correct process so far, would be great, thanks :)

G

 

[Chestermiller]

Why don't you start out by calculating the number of moles of gas in the cylinder using the data for the initial state? Now that you know the number of moles, why don't you use the data for the final state to calculate the volume in the final state? Once you know the initial and final pressures and volumes, you can calculate the polytropic exponent n.

 

[George32]

I did calculate n, I used the ideal gas law to find the volume for the second state and then used logs, but I don't see how that helps find the entropy of the surroundings? I will redo the numbers using your method to check I have the right n.

Thanks for the reply!

I'm using google and "Thermodynamics: An Engineering Approach" by Cengel and Boles, by the way.

 

[Chestermiller]

Suppose you could show that the process was reversible. Could you then determine the change in entropy of the surroundings? What value did you get for n?

 

[George32]

My value for n (polytropic exponent) came out as 1.538 (4s.f.)

If you could show the process were externally reversible, then the entropy generation would have to be 0 and so entropy change of the surroundings would be equal to positive of the entropy change of the system.

The issue is with showing the system is externally reversible. The process is quasi-static and so internally reversible, however this means that the system's temperature increase is smooth and as it ends up 120K above the surroundings, for the system to be totally reversible, no heat could be transferred across that temperature difference. As I understand it.

I know that the net entropy transfer + the entropy generation = change in entropy of the system. So I know one of three variables ( change in the system). I'm required to find the net entropy transfer (assuming that none of the entropy generated contributes to entropy transferred to the surroundings). Finding the entropy generated will answer the last part of the question.


As the surroundings are at constant temperature, I *think* I can say that $$\Delta S = \frac{Qsurr.}{Tsurr.}$$ but now my issue is that I don't know Qsurr. Calculating Qsurr (Taken as the heat transferred to the surroundings) is now the focus of my reading.


After going over my work I noticed a slight miscalculation. My working above wasn't clear and so I was confusing myself. The result I gave for entropy should be specific entropy, so kJ/(kgK)

 

[Chestermiller]

My value for n (polytropic exponent) came out as 1.538 (4s.f.)

If you could show the process were externally reversible, then the entropy generation would have to be 0 and so entropy change of the surroundings would be equal to positive of the entropy change of the system.

The issue is with showing the system is externally reversible. The process is quasi-static and so internally reversible, however this means that the system's temperature increase is smooth and as it ends up 120K above the surroundings, for the system to be totally reversible, no heat could be transferred across that temperature difference. As I understand it.

I know that the net entropy transfer + the entropy generation = change in entropy of the system. So I know one of three variables ( change in the system). I'm required to find the net entropy transfer (assuming that none of the entropy generated contributes to entropy transferred to the surroundings). Finding the entropy generated will answer the last part of the question.


As the surroundings are at constant temperature, I *think* I can say that $$\Delta S = \frac{Qsurr.}{Tsurr.}$$ but now my issue is that I don't know Qsurr. Calculating Qsurr (Taken as the heat transferred to the surroundings) is now the focus of my reading.


After going over my work I noticed a slight miscalculation. My working above wasn't clear and so I was confusing myself. The result I gave for entropy should be specific entropy, so kJ/(kgK)

I'm also not sure what to do with part B. Your equation for ΔS for the surroundings doesn't seem to be correct, because we have no idea what is going on in the surroundings (in addition to heat transfer to or from 20C). But you can calculate Q from the first law. The first step is to calculate the work done on the surroundings (i.e., the PdV work). The second step is to calculate the change in internal energy (dU). Suppose you calculate Q and it comes out positive. What does that tell you about the answer to question c?

This is kind of a hokey problem, because it is hard to understand how the final equilibrium temperature of the system can be 140 C if the surroundings are at 20 C. If the system is in thermal contact with the surroundings, the system at 140 C cannot be an equilibrium state.

 

[George32]

Okie-Dokie.

So I have a result (I think!), let's see what you think. To me it makes sense.

Using the 1st law$$dU=dQ-dW,$$ and the change in energy dU of the surroundings will be equal to that of the system (just in the opposite direction). Compute dW using ∫pdv and p=CV^-n and rearranging to get [tex] \frac{dW=mR(T₂-T₁}{1-n} [/tex] which gives the work done by the system as negative (good so far) 57.09kJ.

Now this is a little more scratchy, but we'll see how it goes. using $$cv=\frac{du}{dT}$$ and separating the variables and integrating to find du as cv(T₂-T₁) we can find dU by then multiplying by the mass (which we have already worked out when finding n, or at least I had) to get dU=46.08kJ.

So for the system we know dU and dW so dQ should be easy enough, I got it as -11.01kJ, which makes sense as that's 11.01kJ of thermal energy being transferred to the surroundings. Then saying that the change in entropy is equal to the heat energy transferred over the temperature, and assuming the temperature of the surroundings stay constant at 20°C, $$\Delta S = \frac{11.01}{293} = 0.03758$$

So then the net change in entropy is the sum of the entropy changes, which is 0.03758 - 0.0313780, which comes out as 0.0061986 kJ/K which is positive but quite small. I therefore took this as being irreversible, as I don't quite know the scale of entropies, as to whether or not this was close enough to 0 to be regarded as reversible.

Thanks for the help Chestermiller!

N.B. I had to use the polytropic exponent n to calculate the work done by the gas, so I managed to get that in, let's hope that was what they meant!

 

[Chestermiller]

Okie-Dokie.

So I have a result (I think!), let's see what you think. To me it makes sense.

Using the 1st law$$dU=dQ-dW,$$ and the change in energy dU of the surroundings will be equal to that of the system (just in the opposite direction). Compute dW using ∫pdv and p=CV^-n and rearranging to get [tex] \frac{dW=mR(T₂-T₁}{1-n} [/tex] which gives the work done by the system as negative (good so far) 57.09kJ.

Now this is a little more scratchy, but we'll see how it goes. using $$cv=\frac{du}{dT}$$ and separating the variables and integrating to find du as cv(T₂-T₁) we can find dU by then multiplying by the mass (which we have already worked out when finding n, or at least I had) to get dU=46.08kJ.

So for the system we know dU and dW so dQ should be easy enough, I got it as -11.01kJ, which makes sense as that's 11.01kJ of thermal energy being transferred to the surroundings. Then saying that the change in entropy is equal to the heat energy transferred over the temperature, and assuming the temperature of the surroundings stay constant at 20°C, $$\Delta S = \frac{11.01}{293} = 0.03758$$

So then the net change in entropy is the sum of the entropy changes, which is 0.03758 - 0.0313780, which comes out as 0.0061986 kJ/K which is positive but quite small. I therefore took this as being irreversible, as I don't quite know the scale of entropies, as to whether or not this was close enough to 0 to be regarded as reversible.

Thanks for the help Chestermiller!

N.B. I had to use the polytropic exponent n to calculate the work done by the gas, so I managed to get that in, let's hope that was what they meant!

There is no reason why the change in internal energy for the surroundings should be equal to minus the change in internal energy for the system. But that isn't really relevant to our discussion.

I checked your values for the work done by the surroundings on the system, and for the change in internal energy of the system, and got the same results. This confirms that the heat removed from the system during the process was 11 kJ. A surroundings at 20 C could accomplish this heat removal. It still isn't clear to me if it is possible to calculate the change in entropy of the surroundings without knowing more details about what is happening in the surroundings (but I doubt it). But the surroundings temperature remaining constant at 20 C while the average temperature of the system was getting higher suggests an irreversible process to me, since it suggests the presence of temperature non-uniformities within the system during the process, which is a clearcut symptom of irreversibility.

Chet

 

[George32]

Yes, sorry, I got myself confused between Total energy and Internal Energy. However the rest should still stand and I'm glad you came to the same conclusions :)

 

[Chestermiller]

Yes. I have an addition to what I said in my previous post. If we regard the surroundings as a separate system, then the Clausius Inequality tells us that the entropy change for the surroundings is greater than Q/T. So we at least know that the entropy change for the surroundings is at the very least, .03758 J/K. This definitely confirms that the process is irreversible. Even though we don't know the exact entropy change of the surroundings, we have still been able to confirm that this particular process is irreversible.

You did very well at working out the details of this problem.

Chet

 

[George32]

Thank you, and thanks for the help! Let's just hope I don't forget before the exams!