CO2 Expansion: Will Temperature Fall Below 0C?

[rancam]

Hi .I'm a beginner at thermodynamics,very limited in my understanding of math.I have a handbook on CO2 properties.According to it the properties for CO2 at 47.3 bar are T=12C,H= -3792,S= -42.6,and for 34.85 bar,
T= 0 C,H= -3339.8,S=-39.33.If the gas expands from 47.3 to 34.85 bar,the gas will liquify because the enthalpy at the pressure of 47.3 is less than that for 34.85 bar.Can you please answer the following question
Will the temperature be 0 C,or will the liquification cause the temperature to fall below 0 C?If so,to determine the temperature would you simply divide the difference between the enthalpies by the heat capacity and subtract the quotient from 0 C (-3339.8 - -3792 =452.2 / 90 = 5.2 C. 0 - 5.2 = -5.2 C.According to the handbook the entropy for CO2 at this temperature (-5.2 C)and pressure (34.85 bar)is -74.64,considerably lower than at the vapor temperature.

 

[Chestermiller]

How are you dropping the pressure: (a) at constant enthalpy (say through a throttling valve) or (b) at constant entropy?

 

[rancam]

It would be at constant enthalpy.

 

[Chestermiller]

It would be at constant enthalpy.

OK. The final temperature is going to be 0 C, and the system is going to be comprised partly of liquid and partly of vapor. It will probably be about 90% vapor, and about 10% liquid. You need to take into account the enthalpy of the liquid and enthalpy of the vapor, and set the total final enthalpy equal to the initial enthalpy. This will enable you to determine the precise proportions of liquid and vapor.

 

[rancam]

Thank you for the help.What I am trying to do is how to use this handbook on CO2 properties in order to determine the entropy of CO2 when it expands from 47.3 bar
and 12 C to 34.85 bar and 0 C.I think I have it figured out.The entropy values for 47.3 bar are -71.82 (liquid).-42.60 (vapor) and 29.22 (vaporisation),and for 34.85 bar
they are;-76.53.-39.33 and 37.20.The final entropy should be -76.53 + 29.22 = -47.31 j / mol.K.Is this correct?If so it would mean that the entropy has gone from -42.6
-47.31,a reduction of 4.71.I thought that in all natural processes entropy must increase.This is a closed system.If I am wrong can you explain how I should be using the
handbook to determine the correct entropy.

 

[Chestermiller]

The entropy change should definitely be positive. I checked on a P-H diagram for CO2 and at all conditions, the entropy increases when the pressure decreases at constant enthalpy. So there must be something wrong with your calculation.

Before we start getting into that, please tell me the units of enthalpy (and entropy) in your posts. You seemed to indicate that the units of s are J/molK, but please check on that. Your enthalpies don't seem to be consistent with your entropies.

I can help you, but only if the data are all consistent in terms of units.

 

[Chestermiller]

OK. I checked the units on your enthalpies and entropies, and they seem to be J/mol and J/mol-K, respectively. So, at 0C, the molar enthalpy of the saturated liquid should be -3340-(37.2)(273.2)=-13503 J/mol. If x is the mole fraction liquid and (1-x) is the mole fraction of vapor at 0 C, then, in order for the enthalpy of the mixture to be the same as the saturated vapor at 12 C, we must have $$-13503x-3340(1-x)=-3792$$

The solution to this equation is x = 0.0445. So the fraction saturated liquid at 0 C is 4.45%, and the fraction saturated vapor is 95.55%.

The entropy in the final state is S = (0.0445)(-76.53)+(0.9555)(-39.33)=-40.98. So the entropy has gone from -42.6 to -41.0, for a change of +1.6 J/mol-K.

 

[rancam]

Thank you for your help.Does this method work for compressing a gas say from 47.3 bar at 12 C to 56.6 bar at 19.5 C?The values I have for liquid enthalpy at

56.6 bar is -11116.5 and for vapor it is -4308.65.So,-11116.5x-4308.65(1-x) = 3792.This gives .076 liq and .924 vap.The entropy values for 56.5 bar are -68.51(liq)

and -45.24 (vap),so .924 x -45.24 = -41.8 +- 5.2 (.076 x -68.51 = -5.2) = -47.Once again the entropy is lower.I must have missused the formula.Can you tell me where I went wrong?

 

[Chestermiller]

Thank you for your help.Does this method work for compressing a gas say from 47.3 bar at 12 C to 56.6 bar at 19.5 C?The values I have for liquid enthalpy at

56.6 bar is -11116.5 and for vapor it is -4308.65.So,-11116.5x-4308.65(1-x) = 3792.This gives .076 liq and .924 vap.The entropy values for 56.5 bar are -68.51(liq)

and -45.24 (vap),so .924 x -45.24 = -41.8 +- 5.2 (.076 x -68.51 = -5.2) = -47.Once again the entropy is lower.I must have missused the formula.Can you tell me where I went wrong?

I don't think that this works for compressing. You can't flow from a lower pressure on one side of a valve or porous plug to a higher pressure at the other side.

 

[rancam]

An undergraduate gave me a similar formula for isentropic compression (S1=S2).S (vapor) for 47.3 bar is -42.6,the S for 56.6 bar is- 45.2 (vapor) and- 68.51 (liquid).
So, -42.6 = (1-q) -68.51+q-45.2,q = 1.111 superheated vapor.Now,is it permissible to do what you did with adiabatic expansion?,namely use the liquid/vapor fraction to
calculate the real entropy.If so then 1.111 x -45.2 = -50.22.This appears illogical,but the S1 = S2 formula is just a method to compute the liquid / vapor fraction,the two pressures could not really have the same entropies.I have read that increasing pressure decreases entropy.Applying the ideal gas equation PV = nRT,should not the superheat increase the pressure (perhaps just for an instant),so for that instant,the real pressure and hence the real entropy would be higher than the values from the
manual (which are at vapor pressure and temperature)?

 

[rancam]

Sorry,in the last sentence I meant to say ,applying the ideal gas equation,the pressure should be higher,and hence the entropy should be lower,than the values at the vapor pressure and temperature (meaning neither subcooled nor superheated).

 

[Chestermiller]

You need to look at a pressure-enthalpy diagram for CO2 to understand better. This is a very valuable type of diagram. There is a p-H diagram for CO2 in Fundamentals of Chemical Engineering Thermodynamics by Moran et al. You can probably also find the diagram online. The diagram even shows constant entropy lines, for your convenience.

If you start out 47.3 bar on the saturation curve, and move along the constant entropy line to higher pressure, the diagram shows that you move into the superheated vapor region. So you only have one phase. For a single phase gas, when you compress it isentropically, both the temperature and the pressure increase. The increase in entropy from the increased temperature exactly cancels the decrease in entropy from the increased pressure, so the entropy stays the same. The ideal gas law applied to an isentropic compression predicts this too, although, in this case, the pressures are a little too high for the ideal gas law to accurately apply.

 

[rancam]

Thanks for your help

 

[rancam]

May I ask one final question?What if CO2 gas at pressures of 47.3 bar, S=-42.6 and 56.6 bar,S= -45.2 were mixed?,as follows.There are two cylinders of the same volume,one open,full of CO2 at 47.3 bar (cylinder A),the other closed (cylinderB).A pipe connects the two cylinders.As a piston pushes the gas in cylinder A through
the pipe,cylinder B opens up so there is no compression.On the way,the gas in the pipe is joined by gas coming from a reservoir,much larger than the cylinders.The pressure of the gas in the reservoir is 56.6 bar.This gas pressurizes the gas in the pipe going from cylinder A to B so that the pressure in cylinder B is 56.5 bar.The question is; what would be the entropy of the gas in cylinder B? Would it be higher than -42.6 or an average of the two?
( - 42.6 + -45.2 / 2 = -43.9 ).My guess is -43.9.My reasoning is,assuming isentropic compression and adiabatic expansion of the two gasses.The lower pressure gas
experiences no expansion and its compression is isentropic,while the higher pressure gas experiences no expansion,so also no change in entropy.So the final entropy would be an average of the two.

 

[Chestermiller]

Sorry, I'm unable to picture what you are describing. Maybe you can provide a diagram(s).

Chet

 

[rancam]

I don't have a picture available at this time.Assuming this scenario is as I have described,do you whether whether the final entropy would be greater than the initial
value of -42.6,or somewhere in between -42.6 and 45.2?

 

[Chestermiller]

As I said, I am unable to figure out the geometry and the scenario. You can draw a diagram in Powerpoint, capture it with the Windows Snipping Tool, save it to your desktop, and upload it rather easily to Physics Forums by hitting the UPLOAD button.

Chet

 

[rancam]

As I am at the public library that might take some time.I was just reading a statement by a member that when gases mix the entropy always increase.My example
is contrived.Lets say the gas does no work on the piston which is activated by an outside force (electrical).I just want to know whether when the lower pressure gas is joined by the higher pressure gas
would the increase in entropy be more than both the higher and lower pressure gas or just the higher pressure?

 

[Chestermiller]

As I am at the public library that might take some time.I was just reading a statement by a member that when gases mix the entropy always increase.My example
is contrived.Lets say the gas does no work on the piston which is activated by an outside force (electrical).I just want to know whether when the lower pressure gas is joined by the higher pressure gas
would the increase in entropy be more than both the higher and lower pressure gas or just the higher pressure?

I don't follow your question, but maybe this will help. Suppose you have an adiabatic chamber with a partition that divides the chamber in half, and you have gas at a high pressure on one side of the partition and the same gas with a lower pressure on the other side of the partition. You poke a hole in the partition and you allow the chamber to re-equilibrate. The final entropy of the system will be higher than original entropy of the system. Does that help?

 

[rancam]

When you say higher than the original entropy of the system do you mean higher than both the higher or lower pressure gas,or just higher than the higher pressure gas.(I am assuming that the higher pressure gas has a lower entropy than the lower pressure gas.Using my example the 47.3 bar gas is -42.6 and the 56.6 bar gas is -45.2).In my example the 47.3 bar gas coming from cylinder A through the pipe is experiencing no expansion and is compressed up to 56.6 by the gas coming from
the high pressure reservoir.It too is experiencing no expansion.The gas flowing through the pipes is traveling at a high velocity so the pressure is much less than 47.3 bar,so there is more space between the molecules.This means the increase in pressure (up to 56.6 bar when the gas enters cylinder B ) is due to the increase in the density of the gas.Would this not mean that this would be an isentropic compression of the 47.3 bar gas,and since there is no expansion of either gas would not this
be isentropic for both gases?If so,would it also imply that the final entropy of the gas in cylinder B would be somewhere between -42.6 and -45.2?

 

[Chestermiller]

When you say higher than the original entropy of the system do you mean higher than both the higher or lower pressure gas,or just higher than the higher pressure gas.

Entropy is an extensive property. What I'm saying is that the final entropy of the combined gases (at the final equilibrium pressure) will be higher than the sum of the original entropies of the higher pressure and lower pressure gases.

 

[Chestermiller]

When you say higher than the original entropy of the system do you mean higher than both the higher or lower pressure gas,or just higher than the higher pressure gas.(I am assuming that the higher pressure gas has a lower entropy than the lower pressure gas.Using my example the 47.3 bar gas is -42.6 and the 56.6 bar gas is -45.2).In my example the 47.3 bar gas coming from cylinder A through the pipe is experiencing no expansion and is compressed up to 56.6 by the gas coming from
the high pressure reservoir.It too is experiencing no expansion.The gas flowing through the pipes is traveling at a high velocity so the pressure is much less than 47.3 bar,so there is more space between the molecules.This means the increase in pressure (up to 56.6 bar when the gas enters cylinder B ) is due to the increase in the density of the gas.Would this not mean that this would be an isentropic compression of the 47.3 bar gas,and since there is no expansion of either gas would not this
be isentropic for both gases?If so,would it also imply that the final entropy of the gas in cylinder B would be somewhere between -42.6 and -45.2?

Unfortunately, you still haven't provided a diagram, so I am at a loss for understanding what this system consists of, and what is happening physically.

Chet

 

[rancam]

Do you mean higher than the average of the sum of the entropies (in my case it would be higher than -42.6+-45.2 =- 87.8 / 2 = -43.9),and is it proportional,so
that if the proportions were 1/4 gas -42.6 and 3/4 at -45.2 the final entropy would be more than -44.55.If so could not the final entropy be less than -42.6?

 

[Chestermiller]

Do you mean higher than the average of the sum of the entropies (in my case it would be higher than -42.6+-45.2 =- 87.8 / 2 = -43.9),and is it proportional,so
that if the proportions were 1/4 gas -42.6 and 3/4 at -45.2 the final entropy would be more than -44.55.If so could not the final entropy be less than -42.6?

Yes, if you are talking about the entropy per unit mass, then the -44.55 would be correct. But, it is better to think of it as S_final>-42.6+3(-45.2) Regarding the final entropy per unit mass being less than -42.6, how could the weighted average of two numbers be less than or equal to either one of them?

 

[rancam]

I was thinking per unit mass.No more questions but a clarification regarding isentropic compression.Using the compression of 47.3 bar isentropically and adiabatically up to 65bar as
an example.Using my manual of CO2 properties the temperature,entropy and enthalpy at 47.3 bar (vapor pressure) is 12C,-42.6, and -3792.At 65 bar (vapor pressure) the temperature is 25.5 C,entropy -48.25 and enthalpy is -5026.4 and density 249.3.At 65 bar and entropy -42.65,the temperature is 35 C.and the enthalpy is -3334.Now,since this is a
closed system would not the increase in temperature (above the 25.5 C at the vapor pressure) to 35 C also increase the pressure (if only for an instant).Would not applying the ideal gas equation (PV = nRT) give an approximate figure for the real pressure?Also I used the figure for the density at the vapor pressure (249.3) and
the higher temperature (35 C) to find the pressure to be about 73 bar.If this is correct,are you saying that this increase in pressure,which would decrease entropy would
be balanced by the increase in entropy resulting from the 9.5 C increase in temperature caused by the work of compression, so the final entropy figure would still be -42.6?

 

[Chestermiller]

I was thinking per unit mass.No more questions but a clarification regarding isentropic compression.Using the compression of 47.3 bar isentropically and adiabatically up to 65bar as
an example.Using my manual of CO2 properties the temperature,entropy and enthalpy at 47.3 bar (vapor pressure) is 12C,-42.6, and -3792.At 65 bar (vapor pressure) the temperature is 25.5 C,entropy -48.25 and enthalpy is -5026.4 and density 249.3.At 65 bar and entropy -42.65,the temperature is 35 C.and the enthalpy is -3334.

You are making this much more complicated than it actually is. You are looking at 3 thermodynamic equilibrium states:

STATE A:
47.3 bars, 12C

STATE B:
65 bars, 25.5 C

STATE C:
65 bars, 35 C

To get from state A to state C, you follow an isentropic compression path. In the end, the entropy in state C is the same as state A.

To get from state B to state C, you hold the pressure constant (by allowing the volume to increase) and gradually add heat to raise the temperature.

Now,since this is a
closed system would not the increase in temperature (above the 25.5 C at the vapor pressure) to 35 C also increase the pressure (if only for an instant).

No. As I said, you are holding the pressure constant by allowing the volume to increase and raising the temperature very gradually.

Would not applying the ideal gas equation (PV = nRT) give an approximate figure for the real pressure?

At these pressures, the ideal gas law is inaccurate.

Also I used the figure for the density at the vapor pressure (249.3) and
the higher temperature (35 C) to find the pressure to be about 73 bar.

As I said, at these pressures the ideal gas law is inaccurate. So 73 bar is not too bad an approximation to 65 bars. If you want to get it more accuarate, you need to use the compressibility factor along with the law of corresponding states, or you need an EOS for CO2 that extends to these pressures.

If this is correct,are you saying that this increase in pressure,which would decrease entropy would
be balanced by the increase in entropy resulting from the 9.5 C increase in temperature caused by the work of compression, so the final entropy figure would still be -42.6?

Basically, this is correct, although the temperature rise from state A to state C is 23 C, not 9.5 C.

But, if you want to calculate the entropy change from state B to state C, it's pretty straightforward, and I can show you how to do that.

 

[rancam]

Yes.Please do.It would be very helpful.

 

[Chestermiller]

Yes.Please do.It would be very helpful.

The change from state B to state C takes place at constant pressure, correct? In terms of Cp, how is the change in enthaply related to the change in temperature at constant pressure? How is the change in enthalpy related to the change in entropy at constant pressure?

 

[rancam]

OK.I'm sorry,I did not bring my CO2 properties manual to the library with me,so I can't look up the Cp at state B and C.I've read it would be an average of the
two.I think your method works like this;First the change in enthalpy; -5026.4 - -3334 = 1692.4.Change in enthalpy is Cp x T so H / Cp = T,or 1692.4 / 178.1 = 9.5.
Change in entropy is Delta Q / abs. T (273 + C),or 1692.4 / 282.5 = 6.If this is a decrease in entropy (-42.6 - 6 = -48.6 ),it is close to the manual's figure of -48.25.Some
of my figures were interpolated so that might account for the difference.If this is so then entropy seems to be the inverse of enthalpy (since the change in enthalpy is positive)divided by absolute temperature, which jives with the definition of unavailable energy.In this case a decrease in entropy means an increase in available energy.
By the way I've just thought of an explanation as to why entropy deceases with pressure.It goes like this.If you have a gas under pressure it must be a closed system.
You can increase the pressure by either adding heat or compressing the gas with a piston (which will both add heat and compress the gas).Because it is a closed system almost all the heat will increase the temperature.A gas that is not under pressure must be an open system and an increase in temperature must come from
adding heat.Because it is open much of this heat is lost to the surroundings in terms of raising the temperature.In other words it takes more heat to raise the temperature for an open system than for a closed one.Using the Clausius equation (which is also the definition of entropy) , Delta Q / abs. T = S ,the Delta Q will
be smaller for a closed system than for an open system (per degree of absolute temperature change),hence the S,or entropy will be less.Is this a correct expanation?

 

[Chestermiller]

OK.I'm sorry,I did not bring my CO2 properties manual to the library with me,so I can't look up the Cp at state B and C.I've read it would be an average of the
two.I think your method works like this;First the change in enthalpy; -5026.4 - -3334 = 1692.4.Change in enthalpy is Cp x T so H / Cp = T,or 1692.4 / 178.1 = 9.5.

The equation I had in mind for constant pressure was:$$dH=C_pdT=TdS$$
Over the temperature interval, the heat capacity is nearly constant, so $C_p=1692.4/9.5=178.1 J/mol$
The entropy change is obtained from $\Delta S=C_p\ln{(308.2/298.7)}=5.58$ J/(mole-K)

If this is so then entropy seems to be the inverse of enthalpy (since the change in enthalpy is positive)divided by absolute temperature, which jives with the definition of unavailable energy.In this case a decrease in entropy means an increase in available energy.

This makes no sense to me, particularly since, entropy is a function of both temperature and pressure, while (at least for an ideal gas) enthalpy is a function only of temperature. Enthalpy and entropy changes need to be evaluated as separate and distinct entities, except in special circumstances.

By the way I've just thought of an explanation as to why entropy deceases with pressure.It goes like this.If you have a gas under pressure it must be a closed system.
You can increase the pressure by either adding heat or compressing the gas with a piston (which will both add heat and compress the gas).Because it is a closed system almost all the heat will increase the temperature.A gas that is not under pressure must be an open system and an increase in temperature must come from
adding heat.Because it is open much of this heat is lost to the surroundings in terms of raising the temperature.In other words it takes more heat to raise the temperature for an open system than for a closed one.Using the Clausius equation (which is also the definition of entropy) , Delta Q / abs. T = S ,the Delta Q will
be smaller for a closed system than for an open system (per degree of absolute temperature change),hence the S,or entropy will be less.Is this a correct expanation?

This makes no sense to me. If the temperature is held constant and you raise the pressure, the volume available to the gas decreases. This reduces the number of states available to the gas, so its entropy decreases.

 

[rancam]

As a beginner I think I've reached the limit of my understanding thermodynamics and its accompanying math.I think its now time to study the required math and perhaps
we can resume this conversation.Thanks for your time and patience.

 

[Chestermiller]

I'd like to make a suggestion. I think it would be very helpful for you to get some practice solving lots of simple problems involving entropy, such as with reversible and irreversible expnasions and compresions of ideal gases and heating/cooling of solids. This should give you a much better feel for how this all plays out.

Chet

 

[rancam]

Thanks again for your help.

 

[Khashishi]

You need to assign units to all your quantities.

 

[Chestermiller]

You need to assign units to all your quantities.

This is a very valid point that rancam needs to pay much more attention to. I was able to decipher the units in post #7, but it would certainly have been much more desirable to carry units through all the posts.

Chet