Compressing steam and water together

[pranj5]

I want to describe here an imaginary experiment. First. take a thick walled cylinder with an open end and pour boiling water into it. The steam coming out will replace the air and then put a piston over the open end and fix it in an airtight way. But, the piston can move up and down.
Now, as the steam will cool to atmospheric level and pressure inside will reduce and the piston will move down. Now, let's move the piston further to compress the steam inside with the remaining water. I want to know what will happen then?
In case, if there is pure steam only inside, the steam will quickly get superheated and began to act like a gas. But, in this case, the steam will always be in contact with water here and have to be saturated and can never act like a gas. I am guessing that in such a case the steam can be compressed with far less energy consumption than pure steam.
How? In this case, the steam will always be saturated and can't be superheated and therefore no need for extra effort to compress superheated steam that will act like a gas.
Am I right?

 

[Chestermiller]

If the piston is frictionless and weightless, the outside pressure will be 1 atm. So as soon as the temperature is dropped below 100C and the system is allowed to equilibrate, all the steam will condense to liquid. The piston will be sitting right on the top of the liquid.

Maybe, instead of allowing the system to cool, you want to consider the case where the system is adiabatic (insulated)?

 

[pranj5]

Right! Now, let's assume some points more. The piston can move freely and frictionless, but we have some control over it so that some space is allowed over the water. And also, there is some hidden mechanism inside the walls of the cylinder that can be switched on and off. In short, we can make the process adiabatic or isothermal at our will.
Now, after the water being cooled, as there is some space available over the water, that should be filled with steam and the pressure of the steam will the same as the saturated steam pressure at that atmospheric temperature level. Now, we have made the system adiabatic and started to move the piston. In short, we started to compress the steam adiabatically. What will happen then?
To be precise, I want to know what will happen if both steam and water are compressed together adiabatically. In such case, the steam will always be in touch with water and therefore can't be superheated and have to remain saturated. In such case, what should be the power consumption?

 

[Chestermiller]

I couldn't understand your first paragraph at all, but the 2nd paragraph seems clear. You have liquid water and water vapor in a cylinder, starting at 100 C and 1 atm. You then compress the system adiabatically and reversibly, and you would like to know the amount of work done. The answer is going to depend on the fraction of the cylinder filled with liquid water to begin with. For example, if there is no liquid water present initially, the answer is going to be very different from the case in which there is say 50% liquid water in the volume initially. Please choose an initial volume fraction of liquid water.

Chet

 

[pranj5]

Do you want to say that amount of power consumption will vary with amount of water. In my opinion, the presence of water will be enough. The water will keep the steam saturated and it can never be superheated. I hope you can understand my point.
And I don't want to mean that the water inside will be at 100C. I clearly want to say that the water inside will be at the same temperature level as the surrounding atmosphere. But, the water will have space over it filled with steam and the temperature of the steam will be the same as the water and pressure of the steam will be the saturated steam pressure at that temperature.
My one and only concern is that whether such an arrangement can reduce the power consumption for steam compression as the steam will always be saturated.

 

[Chestermiller]

Do you want to say that amount of power consumption will vary with amount of water. In my opinion, the presence of water will be enough. The water will keep the steam saturated and it can never be superheated. I hope you can understand my point.
And I don't want to mean that the water inside will be at 100C. I clearly want to say that the water inside will be at the same temperature level as the surrounding atmosphere. But, the water will have space over it filled with steam and the temperature of the steam will be the same as the water and pressure of the steam will be the saturated steam pressure at that temperature.
My one and only concern is that whether such an arrangement can reduce the power consumption for steam compression as the steam will always be saturated.

Suppose there is only one drop of liquid water in the container to start with. Do you still maintain that the drop of water won't evaporate and the system will be saturated throughout the process?

 

[pranj5]

If the process is reversible/idea, then just one single drop will be enough. Whatsoever, we all know that there is nothing irreversible in this universe and hence it's just time wasting to quarrel on whether one drop is enough or two. Let's assume that there is at leas half-a-centimetre of water there. Then?

 

[Chestermiller]

If the process is reversible/idea, then just one single drop will be enough. Whatsoever, we all know that there is nothing irreversible in this universe and hence it's just time wasting to quarrel on whether one drop is enough or two. Let's assume that there is at leas half-a-centimetre of water there. Then?

Just pick a mass percentage of liquid water in the container to begin with. I will then solve the following problem for you, if that is OK:

Insulated Container
Liquid water and water vapor initially in container at equilibrium at 20 C
Adiabatic reversible compression of the contents (no heat exchange with atmosphere outside container).
Calculate change in internal energy, work, and changes in amounts of liquid water and water vapor.

Would you prefer that I solve this, or would you prefer to do the calculation.

 

[pranj5]

I will prefer both. I want to do it in my way and you do it in your own way. Let's see whether that will match or not. In fact, my little knowledge of physics tell me that the power consumption would be just enough to keep the steam saturated while raising both its temperature and pressure.

 

[Chestermiller]

I will prefer both. I want to do it in my way and you do it in your own way. Let's see whether that will match or not. In fact, my little knowledge of physics tell me that the power consumption would be just enough to keep the steam saturated while raising both its temperature and pressure.

OK, but I still need you to specify an initial mass fraction for the liquid water. The pressure vs volume relationship for the contents of the container is going to depend on the initial mass fraction of the liquid water. And this determines the work required per unit mass of cylinder contents.

 

[pranj5]

Kindly tell me how the pressure vs volume relationship will depend on the mass fraction and how much water is necessary for 1 kg of steam (suppose).

 

[Chestermiller]

Kindly tell me how the pressure vs volume relationship will depend on the mass fraction and how much water is necessary for 1 kg of steam (suppose).

You'll see that when we work the problem. Please, just be patient. I am suggesting that we assume 10 mass percent liquid water and 90 mass percent water vapor initially at 20 C. Is that OK with you?

 

[pranj5]

No problem. That means you have taken 900 gm steam and 100 gm water. Right?
So far, this isn't a problem. But, question is, do you consider the steam to be saturated always or not.

 

[Chestermiller]

No problem. That means you have taken 900 gm steam and 100 gm water. Right?

Yes.

So far, this isn't a problem.

Good.

But, question is, do you consider the steam to be saturated always or not.

It depends on how much the system gets compressed (adiabatically and reversibly). Certainly for small volume compression ratios, the system will be saturated.
Before we begin, I have some questions for you too.
1. In this adiabatic reversible compression, do you think that the mass fractions of liquid water and water vapor will (a) remain constant or (b) change.
2. Do you think that, if the proportions change, the amount of work will depend on that?
3. Can you apply the usual ideal gas adiabatic compression equations to the vapor if the number of moles of vapor change?

 

[pranj5]

In reply to your questions:
1. In my opinion, I think the ratio will change. The work done by the piston will change the enthalpy of both the steam and the water and the temperature of water will rise. Therefore, too much water means more power consumption. What is necessary is to limit the amount of water to an optimal level so that most of the work will go to steam to change its enthalpy.
2. Certainly. More water means more work necessary that will add to the as heat.
3. Never! In this case, the steam will remain saturated always and the idea gas compression equations can't be applied here. We just have to calculate the change of enthalpy of steam by subtracting the gross enthalpy of steam at initial stage from the final stage.
In short, the system will be just like a reverse process of multiple cylinder steam engine. Inside multiple cylinder steam engines, steam will go from one cylinder to another (from small diameter to large diameter) and being in the saturated almost always.

 

[Chestermiller]

In reply to your questions:
1. In my opinion, I think the ratio will change. The work done by the piston will change the enthalpy of both the steam and the water and the temperature of water will rise. Therefore, too much water means more power consumption. What is necessary is to limit the amount of water to an optimal level so that most of the work will go to steam to change its enthalpy.
2. Certainly. More water means more work necessary that will add to the as heat.
3. Never! In this case, the steam will remain saturated always and the idea gas compression equations can't be applied here. We just have to calculate the change of enthalpy of steam by subtracting the gross enthalpy of steam at initial stage from the final stage.
In short, the system will be just like a reverse process of multiple cylinder steam engine. Inside multiple cylinder steam engines, steam will go from one cylinder to another (from small diameter to large diameter) and being in the saturated almost always.

Okay. I think we are ready to begin. Do you want to go first or shall I?
Incidentally, in this closed adiabatic system, the work will be equal to the change in internal energy, not the change in enthalpy.

 

[pranj5]

Lets start together. My process is very simple. By this process, as water will always be saturated, therefore the power necessary would be enthalpy of saturated steam at a given lower temperature subtracted from the gross enthalpy at a given higher temperature. As for example, 80C and 100C which we have discussed in another thread. Though the actual consumption will be a little higher just like any process, but unlike other steam compression process, the steam will never get superheated and idea gas laws can't be applied to it.

Incidentally, in this closed adiabatic system, the work will be equal to the change in internal energy, not the change in enthalpy.

I beg to differ. "Internal energy" means only temperature, but enthalpy means both pressure-volume and internal energy. In this process, the pressure and volume wouldn't be the same. Therefore, even in an adiabatic process, it's change in enthalpy, not just internal energy.

 

[Chestermiller]

Lets start together. My process is very simple. By this process, as water will always be saturated, therefore the power necessary would be enthalpy of saturated steam at a given lower temperature subtracted from the gross enthalpy at a given higher temperature. As for example, 80C and 100C which we have discussed in another thread. Though the actual consumption will be a little higher just like any process, but unlike other steam compression process, the steam will never get superheated and idea gas laws can't be applied to it.

I beg to differ. "Internal energy" means only temperature, but enthalpy means both pressure-volume and internal energy. In this process, the pressure and volume wouldn't be the same. Therefore, even in an adiabatic process, it's change in enthalpy, not just internal energy.

I guess we have a fundamental disagreement here. For a closed system, the first law of thermodynamics is given by:
$$\Delta U=Q-W$$
If the process is adiabatic, Q = 0, and we are left with:
$$\Delta U=-W$$
If we are unable to agree on this, we will never be able to agree on the remainder of the solution.

In my judgment, you need to refresh your background on even the most fundamental aspects of thermodynamics. Here is a mini tutorial I prepared for Physics Forums Insights which presents a quick review of the 1st and 2nd laws of thermodynamics. Maybe this will help: https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/

I have obtained a rough estimate of the solution to this problem using the Graphical Method on a pressure enthalpy diagram:

The thick black line in the figure shows the series of states that the cylinder contents passes through as they are compressed. We start off at the lower left end of the line with a saturated mixture of liquid water and water vapor at 20 C and 0.0234 bar, featuring 90 % water vapor and 10% liquid water. As the contents are compressed, the temperature and pressure rise in tandum (according to the saturation vapor pressure relationship) and the liquid water gradually evaporates while the amount of water vapor increases. The total entropy of the contents of the container does not change, because the process is adiabatic and reversible. To satisfy this condition, the amount of liquid water has to be decreasing, while the amount of water vapor has to be increasing. When the contents reaches about 70 C and 0.30 bar, all the liquid will have been evaporated and we will have 100% saturated water vapor in the container. If we compress the contents of the container further than that, the vapor will enter the superheated region. If the contents were compressed to 1 bar, the superheated vapor will be at a temperature close to 200 C.

I will fine tune these results using the data in the steam tables, if your are interested.

Chet

 

[pranj5]

From the perspective of !st law of thermodynamics, ΔU means gross enthalpy of the system. It doesn't denote only the internal energy. Can you explain why the work done is added only to the internal energy of the system only if both pressure, volume and temperature has been changed?
Wikipedia (https://en.wikipedia.org/wiki/Enthalpy) says something different and what you have said above is true is saturated steam behaves like an ideal gas. But, that isn't the fact here.

 

[Chestermiller]

From the perspective of !st law of thermodynamics, ΔU means gross enthalpy of the system. It doesn't denote only the internal energy. Can you explain why the work done is added only to the internal energy of the system only if both pressure, volume and temperature has been changed?
Wikipedia (https://en.wikipedia.org/wiki/Enthalpy) says something different and what you have said above is true is saturated steam behaves like an ideal gas. But, that isn't the fact here.

I stand by everything I said. Please see my private conversation comments where I feel like I can speak more frankly.

Chet

 

[pranj5]

The hot steam will stay over the water and what it can do maximum is to raise the temperature of the upper layer of water. As far as I know, water (or any liquid will evaporate) when pressure over them will reduce. How the water can evaporate when the pressure over them increase and the temperature is always the saturation temperature. As far as I know, at saturation temperature you have give heat from outside to make steam from water.

 

[Chestermiller]

The hot steam will stay over the water and what it can do maximum is to raise the temperature of the upper layer of water. As far as I know, water (or any liquid will evaporate) when pressure over them will reduce. How the water can evaporate when the pressure over them increase and the temperature is always the saturation temperature. As far as I know, at saturation temperature you have give heat from outside to make steam from water.

This is not correct. When you compress a vapor adiabatically, its temperature increases. This increase in temperature of the vapor can cause the liquid to be heated. Heating the liquid can cause its temperature to rise and can also cause some of it to evaporate.

In the calculations that we are talking about, we are assuming that the combination of liquid and gas in the cylinder is being compressed very slowly, so that, at any time during the process, part of the heat from compressing the gas is transferred to the liquid, such that the liquid and vapor are essentially (nearly) at the same temperature at all times. By compressing the system adiabatically and reversibly (i.e., slowly), we are obtaining the minimum possible amount of work required to compress the contents. If we were to compress the contents very rapidly (irreversibly), we would be generating extra heat in the system, and the amount of work would have to be greater (depending on how fast we compressed it). That's why we are looking at the best, most optimum, case of very slow reversible compression.

In post #18, I showed how the mass fraction of water vapor increases in the system as we raise the pressure (compress the contents) under saturation conditions. This continues until all the liquid has evaporated. From that point on, if we raise the pressure any more, we will be compressing superheated vapor. What I haven't shown yet is how the combined volume of liquid and vapor changes as we raise the pressure or how much work is done at various pressures per kg of water. Are you interested in seeing this? (I would think you would, considering the comparison you are trying to make).

Chet

 

[pranj5]

I certainly would. Whatsoever, kindly think about the process in opposite way. That means the piston will move in the opposite direction. In that case, the water will evaporate and steam will fill up the vacuum. The latent heat of vaporisation of the steam will come from the water and it will become colder and at the end will become ice.
What we are discussing is if the piston moves inward. Simply common sense says that what will happen is just opposite. Steam will be compressed and heated and a little part of it will add to the water to raise its temperature. In short, the amount of water will increase instead of decreasing at the expense of steam. And at 220 bars of pressure and 374C temperature i.e. at the triple point, both water and steam will mix up altogether.
Kindly note that until and unless there is water inside the cylinder, the compression will follow the saturation curve. In short, the work done will be added to the gross enthalpy of steam and the steam will always be in saturated state.
In fact, your long experience working in industries is now forcing you to a particular direction. You have seen that during steam compression, huge amount of energy is necessary and you have started from that point. But, I have doubt whether you have worked on any experiment similar to what we are discussing here. At present, almost all the steam compressor companies work with steam at pressure level higher than atmosphere and as far as I know, no company is working on compressing steam from very low pressure level and how to reduce power consumption on that matter.

 

[Chestermiller]

I certainly would. Whatsoever, kindly think about the process in opposite way. That means the piston will move in the opposite direction. In that case, the water will evaporate and steam will fill up the vacuum. The latent heat of vaporisation of the steam will come from the water and it will become colder and at the end will become ice.

Actually, this is not exactly what would happen. If the vapor were allowed to expand adiabatically and reversibly against a piston, the vapor would be doing work. So its temperature would have to drop. This would do two things: 1. cause the liquid temperature to also drop and 2. cause some of the vapor to condense to form more liquid. So the condensation of the vapor would have a modulating effect on the temperature drop of the combined system. The decrease in temperature of the liquid would be the same as the decrease in temperature of the vapor, and the system would remain saturated. The cooling would continue until the temperature of the system reached 0 C and ice began to form. So it is not evaporation that is causing the temperature of the system to fall, but rather expansion cooling of the vapor. This is at least the case in our system where the mass of vapor far exceeds the mass of liquid. It might not be the case however in a situation where the mass of liquid were on the same order as the mass of vapor.

What we are discussing is if the piston moves inward. Simply common sense says that what will happen is just opposite.

Yes, but not as you described it. It would be as I described it.

Steam will be compressed and heated and a little part of it will add to the water to raise its temperature.

Yes and part of it will go into evaporating some of the water to form more vapor.

In short, the amount of water will increase instead of decreasing at the expense of steam. And at 220 bars of pressure and 374C temperature i.e. at the triple point, both water and steam will mix up altogether.

Actually, for our system, the amount of liquid water would decrease to produce more water vapor. The compression will follow the saturation curve until all the water is evaporated. This is all fully consistent with the constant-entropy operating curve I drew in the figure. This curve represents saturated conditions until the temperature reaches about 65 C, and all the water has evaporated.

Kindly note that until and unless there is water inside the cylinder, the compression will follow the saturation curve. In short, the work done will be added to the gross enthalpy of steam and the steam will always be in saturated state.

You can't really think this for the limiting case in which there is only one drop of liquid water initially in a huge excess of water vapor. After all, if that drop of liquid were not there, the vapor would immediately compress into the superheated region.

In fact, your long experience working in industries is now forcing you to a particular direction. You have seen that during steam compression, huge amount of energy is necessary and you have started from that point. But, I have doubt whether you have worked on any experiment similar to what we are discussing here. At present, almost all the steam compressor companies work with steam at pressure level higher than atmosphere and as far as I know, no company is working on compressing steam from very low pressure level and how to reduce power consumption on that matter.

Please don't tell me what I am thinking and that I am biased by my prior experience. You can't possibly know that. All I am doing is letting the physics of the problem and the mathematics lead me in the direction that it wants to lead me. I am just following what it is telling me.

 

[pranj5]

Actually, this is not exactly what would happen. If the vapor were allowed to expand adiabatically and reversibly against a piston, the vapor would be doing work. So its temperature would have to drop. This would do two things: 1. cause the liquid temperature to also drop and 2. cause some of the vapor to condense to form more liquid. So the condensation of the vapor would have a modulating effect on the temperature drop of the combined system. The decrease in temperature of the liquid would be the same as the decrease in temperature of the vapor, and the system would remain saturated. The cooling would continue until the temperature of the system reached 0 C and ice began to form. So it is not evaporation that is causing the temperature of the system to fall, but rather expansion cooling of the vapor. This is at least the case in our system where the mass of vapor far exceeds the mass of liquid. It might not be the case however in a situation where the mass of liquid were on the same order as the mass of vapor.

What I want to mean is that if the piston has been forcefully drawn outwards. That doesn't mean that the steam has to do work on the piston. That simply means expanding the vacuum and fulfilling that with evaporated steam. To do some work, the pressure of the steam should be more than the pressure over the piston on the other side. But, in this case, the steam pressure is far less than outside pressure. In such a scenario, kindly tell me how you can be correct.

Yes, but not as you described it. It would be as I described it.

How? Like adiabatic compression, adiabatic expansion can also be possible

Yes and part of it will go into evaporating some of the water to form more vapor.

How? The steam will be saturated and part of it will go water to raise its temperature to the same level as the steam. Evaporation will occur, when the temperature of water will be higher than the steam.

Actually, for our system, the amount of liquid water would decrease to produce more water vapor. The compression will follow the saturation curve until all the water is evaporated. This is all fully consistent with the constant-entropy operating curve I drew in the figure. This curve represents saturated conditions until the temperature reaches about 65 C, and all the water has evaporated.

Again, how? Until and unless there is water, the steam will be saturated and evaporation can only occur when the temperature of water will be higher than the temperature of steam. But, that's simply beyond common sense.

You can't really think this for the limiting case in which there is only one drop of liquid water initially in a huge excess of water vapor. After all, if that drop of liquid were not there, the vapor would immediately compress into the superheated region.

If the vapour was dry, saturated; then and then only it would be possible. In reality, you can't get dry, saturated steam alone without a little superheating. The excess heat will quickly suck the water drop. But, if there is some reasonable amount of water there, that means some will evaporate to keep the steam in saturated condition. In theory, if the steam is EXACTLY saturated, then this single drop would be enough. But, in reality you can't get exactly saturated steam and that's why talking about having just a drop of water is foolish.

Please don't tell me what I am thinking and that I am biased by my prior experience. You can't possibly know that. All I am doing is letting the physics of the problem and the mathematics lead me in the direction that it wants to lead me. I am just following what it is telling me.

You haven't pointed out that at which point I have gone beyond physics.
What I know is something with common sense. As for example, if you go from point A to point B and return to point A, that means the conditions are unchanged in ideal case. Let's take something like this in our experiment. First, expand the steam by moving the piston outwards by some level. That means decrease in temperature and pressure of the both steam and water and both will be at saturated stated as they are in contact with each other. In the next step, the piston has been returned back to its original position. That means increase in temperature and pressure of steam. If in the first case, water will be evaporated then certainly in the opposite direction, steam will be condensed back and no evaporation will take place.
I hope I am right in this case.

 

[Chestermiller]

What I want to mean is that if the piston has been forcefully drawn outwards. That doesn't mean that the steam has to do work on the piston. That simply means expanding the vacuum and fulfilling that with evaporated steam. To do some work, the pressure of the steam should be more than the pressure over the piston on the other side. But, in this case, the steam pressure is far less than outside pressure. In such a scenario, kindly tell me how you can be correct.

If the piston is drawn forcibly, then the process is not reversible. However, the steam is still doing work on the piston, even when you draw it forcibly. On one side of the piston is atmospheric pressure times the piston area minus the force that you apply. On the other side of the piston is the steam pressure at the piston face times the piston area. Assuming that the piston is frictionless and has very little mass, the steam pressure times the piston area must match the atmospheric pressure times the piston area minus the force you apply: $p_vA=p_aA-F$. So the vapor is still doing work on the piston.

If you do carry out this operation by forcefully withdrawing the piston, the expansion is adiabatic, but irreversible, and that is very different from an adiabatic reversible expansion. In this case, you have viscous dissipation of mechanical energy to additional heat, which results in an increase in entropy; and the amount of condensation can be less, and even allow evaporation. But it will then not be the opposite of adiabatic irreversible compression. Adiabatic reversible compression will also result in an increase in entropy of the system; and, in this case, the amount of liquid evaporation will be even greater than if the compression is carried out reversibly.

If you would like to look at some cases of adiabatic irreversible compression, I can help you analyze that. But, you will not be happy with the results.

How? Like adiabatic compression, adiabatic expansion can also be possible

Yes. But, as I described it, adiabatic reversible expansion will result in an increase in liquid water, not a decrease. And adiabatic expansion will result in a decrease in liquid water, not and increase.

How? The steam will be saturated and part of it will go water to raise its temperature to the same level as the steam. Evaporation will occur, when the temperature of water will be higher than the steam.

Compressing the vapor causes its temperature through the adiabatic reversible process to rise slightly above that of the liquid during. In this limiting situation, the liquid temperature lags slightly behind the liquid temperature. This allows heat to be transferred from the vapor to the liquid to both raise its temperature and cause some evaporation.

Again, how? Until and unless there is water, the steam will be saturated and evaporation can only occur when the temperature of water will be higher than the temperature of steam. But, that's simply beyond common sense.

Here are your own words (which I agree with): But, if there is some reasonable amount of water there, that means some will evaporate to keep the steam in saturated condition. I hope this answers your question.

If the vapour was dry, saturated; then and then only it would be possible. In reality, you can't get dry, saturated steam alone without a little superheating. The excess heat will quickly suck the water drop. But, if there is some reasonable amount of water there, that means some will evaporate to keep the steam in saturated condition. In theory, if the steam is EXACTLY saturated, then this single drop would be enough. But, in reality you can't get exactly saturated steam and that's why talking about having just a drop of water is foolish.

Now that you have admitted that one drop evaporates, that destroys the whole remainder of your arguments. Then you correctly transitioned to "some reasonable amount of water" and reached the correct conclusion: But, if there is some reasonable amount of water there, that means some will evaporate to keep the steam in saturated condition.

This is what I have been saying.

What I know is something with common sense. As for example, if you go from point A to point B and return to point A, that means the conditions are unchanged in ideal case. Let's take something like this in our experiment. First, expand the steam by moving the piston outwards by some level. That means decrease in temperature and pressure of the both steam and water and both will be at saturated stated as they are in contact with each other. In the next step, the piston has been returned back to its original position. That means increase in temperature and pressure of steam.

I totally agree with all this if the expansion and compression are adiabatic and reversible.

If in the first case, water will be evaporated then certainly in the opposite direction, steam will be condensed back and no evaporation will take place.
I hope I am right in this case.

For the relative amounts of liquid water and water vapor that we have agreed upon, I disagree with this statement. It should be:
If in the first case, water will be condensed then certainly in the opposite direction, steam will be evaporated back and no condensation will take place.
Again, here are your own words: "But, if there is some reasonable amount of water there (during compression), that means some will evaporate to keep the steam in saturated condition."

 

[pranj5]

If the piston is drawn forcibly, then the process is not reversible. However, the steam is still doing work on the piston, even when you draw it forcibly. On one side of the piston is atmospheric pressure times the piston area minus the force that you apply. On the other side of the piston is the steam pressure at the piston face times the piston area. Assuming that the piston is frictionless and has very little mass, the steam pressure times the piston area must match the atmospheric pressure times the piston area minus the force you apply: $p_vA=p_aA-F$. So the vapor is still doing work on the piston.

If you do carry out this operation by forcefully withdrawing the piston, the expansion is adiabatic, but irreversible, and that is very different from an adiabatic reversible expansion. In this case, you have viscous dissipation of mechanical energy to additional heat, which results in an increase in entropy; and the amount of condensation can be less, and even allow evaporation. But it will then not be the opposite of adiabatic irreversible compression. Adiabatic reversible compression will also result in an increase in entropy of the system; and, in this case, the amount of liquid evaporation will be even greater than if the compression is carried out reversibly.

If you would like to look at some cases of adiabatic irreversible compression, I can help you analyze that. But, you will not be happy with the results.

Why not? If compression can be reversible, why not expansion. As per you, the steam will do some work. But to do work, you have to move against a force (in this case atmospheric pressure) and your force must be greater than that. If what you are saying is true, then steam wouldn't form, but rather condense inside Evaporators. The steam pressure here can't do any work as it can't overcome atmospheric pressure. The work done here is only by the external force that displaces the piston.

Yes. But, as I described it, adiabatic reversible expansion will result in an increase in liquid water, not a decrease. And adiabatic expansion will result in a decrease in liquid water, not and increase.

Your words are confusing here.

Compressing the vapor causes its temperature through the adiabatic reversible process to rise slightly above that of the liquid during. In this limiting situation, the liquid temperature lags slightly behind the liquid temperature. This allows heat to be transferred from the vapor to the liquid to both raise its temperature and cause some evaporation.

The temperature of liquid water will rise too but as the steam is saturated and therefore the temperature of water can't rise above the saturation level and therefore no vaporisation will occur.

Here are your own words (which I agree with): But, if there is some reasonable amount of water there, that means some will evaporate to keep the steam in saturated condition. I hope this answers your question.

That's the reality! In ideal case, a single drop will be enough. The same amount of water will be evaporated to drag the temperature steam from superheated to saturated level and after that, no evaporation will occur.

Now that you have admitted that one drop evaporates, that destroys the whole remainder of your arguments. Then you correctly transitioned to "some reasonable amount of water" and reached the correct conclusion: But, if there is some reasonable amount of water there, that means some will evaporate to keep the steam in saturated condition.
This is what I have been saying.

Again, I want to say that this is in reality, where the initial steam will be a little superheated.

For the relative amounts of liquid water and water vapor that we have agreed upon, I disagree with this statement. It should be:
If in the first case, water will be condensed then certainly in the opposite direction, steam will be evaporated back and no condensation will take place.
Again, here are your own words: "But, if there is some reasonable amount of water there (during compression), that means some will evaporate to keep the steam in saturated condition."

How can water be condensed when pressure is reduced over it?

 

[Chestermiller]

3-Stage Pressure Increment

Let's do the equilibration at each incremental pressure change in 3 stages (rather than all at one time). First we increase the pressure of the vapor by a small $\Delta p$. This causes its temperature to increase by a small $\Delta T$, and reach a new higher temperature $T_1$ (while the liquid is still at the original temperature that existed at the beginning of the pressure increment $T_0$). Then, for the 2nd stage, at the constant pressure, we allow the vapor and liquid to equilibrate thermally without allowing any liquid to evaporate or vapor to condense. So the temperature at the end of this stage will be $T_2$, where $T_0<T_2<T_1$. Now, the temperatures and pressures of the liquid and the vapor at the end of this stage will both be the same. But, the question is, what is there to guarantee that this new combination of temperatures and pressures will correspond to a saturation condition? The answer is nothing. If the new pressure is lower than the equilibrium vapor pressure at the new system temperature $T_2$, then some liquid will have to evaporate such that the system the system temperature will drop a little. Now the temperature and pressure of the liquid and vapor will be the same, and vapor-liquid equilibrium will have been reestablished at a new temperature and pressure.

Here is a rough calculation to illustrate this. We have 0.1 kg of liquid water and 0.9 kg of water vapor at 20 C and 0.0234 bars. We insert a thin massless, frictionless, insulated barrier between the vapor and liquid to begin with, and we compress the vapor adiabatically and reversibly to twice the pressure, 0.0468 bars. Because of the insulated barrier, the temperature of the liquid won't change, but the temperature of the vapor will (approximately) increase according to the adiabatic reversible compression equation for an ideal gas, given by:$$\frac{T_1}{T_0}=\left(\frac{0.0468}{0.0234}\right)^{\frac{\gamma -1}{\gamma}}$$This gives $$T_1=348K=75C$$. The liquid is still at 20 C.

We now replace the massless, frictionless, insulated barrier with a massless, frictionless, diabatic (conductive) barrier that allows the liquid and vapor to thermally equilibrate at the new pressure. The heat capacity of the liquid water is about 4.18 kJ/kg and the heat capacity of the vapor is about 2.00 kJ/kg. If we make the assumption that no liquid evaporates and no vapor condenses and allow the system to equilibrate thermally, the new equilibrated temperature of the system will be given by the equation:
$$(0.9)(2.00)(T_2-75)+(0.1)(4.18)(T_2-20)=0$$
The solution to this equation for the equilibrated temperature $T_2$ (again, assuming there was no evaporation or condensation) is:
$$T_2=64.6C$$
At 64.6C, the equilibrium vapor pressure of water is approximately 0.26 bars. This is way higher than the 0.0468 bars which represents the system pressure. The temperature at which the equilibrium vapor pressure is 0.0468 bars is only about 30 C. So some of the liquid would have to evaporate to cool the system down to the actual applied pressure. The heat of vaporization of water in this range is about 2260 kJ/kg. About 0.03 kg of liquid water would have to evaporate to cool the system back down to this temperature. So the final amounts of liquid water and later vapor would be about 0.07 kg and 0.93 kg respectively.

 

[Chestermiller]

Even if you don't agree with anything I have said so far, can we at least agree on the following two points:

1. For an adiabatic reversible process applied to a closed system, the change in entropy of the system is equal to zero.

2. The Saturated Steam Tables give accurate values for the following thermodynamics properties of liquid water and water vapor under saturation (temperature and pressure) conditions: internal energy per unit mass, enthalpy per unit mass, entropy per unit mass.

What do you think? Do you agree with these two items?

 

[Chestermiller]

I have obtained an interesting analytic solution to this problem which I think that others (e.g., @mech-Engineer) besides @pranj5 (who will probably not agree with what I've done here) will probably find cool.

The only assumption in the analysis is that the effect of temperature and pressure on the specific volume and heat capacity of the liquid water is negligible. No assumption is made about the vapor being an ideal gas.

For the adiabatic reversible compression of the combination of liquid water and water vapor inside the cylinder, the first law of thermodynamics tells us that: $$dU=-PdV\tag{1}$$or equivalently $$dH=VdP\tag{2}$$
where U is the internal energy of the cylinder contents, H is the enthalpy, P is the pressure, and V is the volume. In the present development, we will find it more convenient to use Eqn. 2, involving the enthalpy.

The enthalpy of the cylinder contents at temperature T and pressure P is given by:
$$H=(n_L+n_V)C(T-T_R)+(n_L+n_V)v_L(P-P_R)+n_V\lambda (T)\tag{3}$$
where n_L and n_V are the number of moles of liquid water and water vapor respectfully, C is the heat capacity of liquid water, T_R and P_R are the temperature and pressure in the reference state of zero enthalpy (liquid water at 0C and 1 atm), v_L is the specific volume of liquid water, and $\lambda (T)$ is the heat of vaporization of water at temperature T.

The volume of the liquid water and the water vapor in the container at temperature T and pressure P are given by:
$$V=n_Lv_L+n_Vv_V=(n_L+n_V)v_L+n_V(v_V-v_L)\tag{4}$$where v_V is the specific volume of the vapor. It is important to note that, in Eqns. 3 and 4, the total number of moles of liquid and vapor in the cylinder $(n_L+n_V)$ is a constant. If we substitute Eqns. 3 and 4 into Eqn. 2, we obtain:
$$(n_L+n_V)CdT+d(n_V\lambda (T))=n_V(v_V-v_L)dP\tag{5}$$
As long as there is still liquid remaining in the cylinder during this adiabatic reversible process, the vapor and liquid will be at saturation conditions at all times. The relationship between pressure and temperature under saturation conditions is given by the Clapeyron equation:
$$\frac{dP}{dT}=\frac{\lambda (T)}{T(v_V-v_L)}\tag{6}$$
If we substitute Eqn. 6 into Eqn. 5, we obtain:
$$(n_L+n_V)CdT+d(n_V\lambda(T))=\frac{n_V\lambda(T)}{T}dT\tag{7}$$
This equation can be simplified mathematically to:$$C\frac{dT}{T}+d\left(\frac{x_V\lambda(T)}{T}\right)=0\tag{8}$$
where x_V is the mass fraction of vapor in the cylinder at any time:
$$x_V=\frac{n_V}{(n_L+n_V)}$$
The left hand side of Eqn. 8 is an exact differential, and Eqn. 8 can be rewritten as:
$$d\left(C\ln{T}+\frac{x_V\lambda (T)}{T}\right)=0\tag{9}$$
If we subtract $d(C\ln T_R)$ from both sides of Eqn. 9, we obtain:
$$d\left(C\ln{(T/T_R)}+\frac{x_V\lambda (T)}{T}\right)=0\tag{10}$$
The term in parenthesis in Eqn. 10 represents the entropy per unit mass of a saturated mixture of liquid water and water vapor at mass fraction vapor x_V and temperature T, relative to the reference state of purely liquid water at 0 C (T_R=273 K): $$s(x_V,T)=C\ln{(T/T_R)}+\frac{x_V\lambda (T)}{T}\tag{11}$$
If we integrate Eqn. 10 between the initial state of our liquid/vapor mixture $(x_{V0},T_0)$ and some other compressed state during our adiabatic reversible compression, we obtain:

$$s(x_V,T)=C\ln{(T/T_R)}+\frac{x_V\lambda (T)}{T}=s(x_{V0},T_0)=C\ln{(T_0/T_R)}+\frac{x_{V0}\lambda (T_0)}{T_0}\tag{12}$$

We can check the accuracy of the present analysis by calculating the entropy of the system in the initial state $x_V=0.9$ and $T_0=293K$ from Eqn. 12 and comparing the results with the corresponding entropy value determined from the steam tables. Implementing this, we have:
$$s(0.9,293)=4.179\ln{(293/273)}+\frac{(0.9)(2454)}{293}=7.833$$
The corresponding value calculated from the steam tables is 7.830

 

[Mech_Engineer]

I think it's time we enshrine this thread and print it out as a textbook

Fantastic posts Chet!

 

[pranj5]

You have done a good job. But, there is a flaw at the starting.
As per you "we allow the vapour and liquid to equilibrate thermally without allowing any liquid to evaporate or vapour to condense". If the system is closed i.e. no heat can enter or leave the system, that's simply impossible. The vapour will be at a higher temperature and water will be at a lower temperature. To equilibrate, heat should enter to the water from steam and as both are in saturated condition, that means liquefaction of a small amount of steam.
And another point is that, during compression you have considered the steam to be superheated and therefore act like an ideal gas. That's the biggest point of objection for me. Steam in contact with water can never get superheated and always remain in saturated state. It's a basic property of steam.
We simply can't reach to a proper solution as long as we consider steam to be superheated and act like an ideal gas, we will consider far higher power consumption and can't differentiate when water and steam are compressed together in comparison to when steam is compressed alone.

 

[pranj5]

Even if you don't agree with anything I have said so far, can we at least agree on the following two points:

1. For an adiabatic reversible process applied to a closed system, the change in entropy of the system is equal to zero.

2. The Saturated Steam Tables give accurate values for the following thermodynamics properties of liquid water and water vapor under saturation (temperature and pressure) conditions: internal energy per unit mass, enthalpy per unit mass, entropy per unit mass.

What do you think? Do you agree with these two items?

I want to start from those points from the very beginning.

 

[pranj5]

In the equation 3, I want to say that

H=(nL+nV)C(T−TR)+(nL+nV)vL(P−PR)+nVλ(T)

PR is the saturated steam pressure at 20C, while P is atmospheric pressure. That means it's a negative quantity. In the same equation, the part

(nL+nV)C(T−TR)

is also doubtful because specific heat of water and specific heat of steam isn't same.
At the same equation, the part nVλ(T) is also doubtful because λ(T) usually has been given in cal/gm or J/gm, not in moles.

 

[Chestermiller]

In the equation 3, I want to say that PR is the saturated steam pressure at 20C, while P is atmospheric pressure. That means it's a negative quantity. In the same equation, the part is also doubtful because specific heat of water and specific heat of steam isn't same.
At the same equation, the part nVλ(T) is also doubtful because λ(T) usually has been given in cal/gm or J/gm, not in moles.

It looks like we are finally starting to get serious about this problem. You seem to be comfortable with what I have said in Post #30, at least up through Eqn. 2. This is great because, if we continue through this analysis together, I can tell you for certain at this point that you are now "hooked."

Regarding Eqn. 3. You seem to misunderstand how I obtained the change in enthalpy H relative to the reference state I have selected. My goal is to determine the enthalpy of the combination of $n_L$ moles of liquid water and $n_V$ moles of water vapor in a saturated mixture at temperature T and equilibrium vapor pressure P, relative to the reference state of $(n_L+n_V)$ moles of purely liquid water at the reference state of 0 C and 1 atm. So the initial and final states that I am looking at are:

State 1:
$(n_L+n_V)$ moles of liquid water
Temperature = $T_R$ = 0 C
Pressure = $P_R$ = 1 atm.

State 2:
$n_L$ moles of liquid water
$n_V$ moles of water vapor
Temperature = T
Pressure = equilibrium vapor pressure of water at temperature T

State 2 is one of the states that can exist in our actual system, but State 1 cannot. Of course, since enthalpy is only a function of state, it doesn't matter what process we apply to evaluate the enthalpy in State 2 relative to reference State 1. And in the end, the parameters related to State 1 will all drop out of our analysis.

Here is the 3 step process I have I devised for getting the enthalpy of State 2 relative to State 1.

I start out with $(n_L+n_V)$ moles of liquid water at 0 C inside a vertical cylinder with a piston sitting on top of the liquid water (and vacuum surrounding the cylinder and piston). There are a set of weights sitting on top of the piston, some of which can be removed to decrease the pressure on the liquid within the cylinder. There is no vapor in the cylinder to begin with, and the total pressure exerted by the piston and weights on the liquid is 1 atm. Thus, the system is in State 1.

Step 1: I add heat to the cylinder contents to raise their temperature to temperature T (< 100 C) without removing weights from the piston, so that the total pressure is still 1 atm and the contents remain a liquid. So the total number of moles $(n_L+n_V)$ is heated as a liquid.

Step 2: Now that I am at temperature T, I remove just enough weights from the piston to drop the total pressure on the liquid (assumed incompressible) from 1 atm. to the equilibrium saturation vapor pressure P of water at temperature T.

Step 3: Now that I am at the equilibrium vapor pressure at temperature T, I add heat to the cylinder (without removing any more weights from the piston) until $n_V$ moles of liquid have evaporated. During this change, the temperature is constant at T.

At the end of Step 3, I have arrived at State 2.

Now, I'm going to let you work out the total change in enthalpy H in going from State 1 to State 2.

Step 1: If C is the heat capacity of liquid water at constant pressure, what is the change in enthalpy of the $(n_L+n_V)$ moles of liquid water in going from temperature $T_R$ to temperature T?

Step 2: If $v_L$ is the volume per mole of liquid water, what is the change in enthalpy of the $(n_L+n_V)$ moles of liquid water in going from the pressure $P_R$ to the equilibrium vapor pressure P at constant temperature?

Step 3: If $\lambda (T)$ represents the heat of vaporization per mole of saturated liquid water to saturated water vapor at temperature T and equilibrium vapor pressure P, what is the change in enthalpy of the $n_V$ moles of water that vaporize in Step 3?

What is the total change in enthalpy for Steps 1-3?

We can continue after you have completed this.