Compressing steam and water together

[pranj5]

I just want to know, where the decreased enthalpy will go. Against post 44, I have asked just one question which isn't answered yet. At least I can say that if pressure is reduced over the water, the change in enthalpy of the water is negative.

 

[Chestermiller]

I just want to know, where the decreased enthalpy will go.

You yourself just indicated that, for the transition from State A to State B, the change in enthalpy per mole is $\Delta h=v\Delta P$ (negative). So you already know that the result in post #44 is correct. Now you are asking where the decreased enthalpy went to. The answer is that, because enthalpy is merely a mathematically defined quantity, it is not, according to the first law, something that is required to be accounted for. The focus of the first law for a closed system is the internal energy, not the enthalpy. This is fundamental, and is presented in every thermodynamics book: the change in internal energy of a system $\Delta U$ is equal to the heat added Q minus the work done on the surroundings W. So, when we calculate that the enthalpy change is negative when the pressure on an incompressible material decreases adiabatically, the $v\Delta P$ does not have to be accounted for, and is just a mathematical term that is carried along for completeness. However, we find that in our problem that , when we substitute the equation for the enthalpy and the volume (Eqns. 3 and 4) into Eqn. 2, the term in question exactly cancels out with another term on the right hand side of the equation. So, in the end, the term in question has absolutely no effect on the results.

 

[pranj5]

Can't agree to you on that point. As per wikipedia (https://en.wikipedia.org/wiki/Enthalpy), enthalpy is a measurement of energy in a thermodynamic system. Therefore, it has something to do with the first law of thermodynamics. It seems that you are mixing up total energy with internal energy. The first law is applicable not only to internal but also on total energy.

 

[Chestermiller]

Can't agree to you on that point. As per wikipedia (https://en.wikipedia.org/wiki/Enthalpy), enthalpy is a measurement of energy in a thermodynamic system. Therefore, it has something to do with the first law of thermodynamics. It seems that you are mixing up total energy with internal energy. The first law is applicable not only to internal but also on total energy.

OK. Since you are not able to accept my explanation of the tiny $v\Delta P$ term for Step 2 as just a bookkeeping entry, and without my being able to provide you with a physical interpretation of this term that satisfies your intuition, I am proposing to re-solve the problem solely in terms of internal energy U. Certainly, any problem that can be solved in terms of enthalpy can equally well be solved in terms of internal energy, right? Is this acceptable to you? We will not be saying the word enthalpy again or using the enthalpy function again. How does that grab you?

 

[pranj5]

Not at all. Enthalpy is something more than just internal energy and it's very important factor in this whole process. Actually, I have started this thread to know whether steam will be in saturated state or not during compression with water and without considering, it just can't be solved.

 

[Chestermiller]

Not at all. Enthalpy is something more than just internal energy and it's very important factor in this whole process. Actually, I have started this thread to know whether steam will be in saturated state or not during compression with water and without considering, it just can't be solved.

So you are saying that this problem can't be solved solely in terms of internal energy, and one can only solve it only in terms of enthalpy. Is that your engineering judgment?

 

[pranj5]

That's my judgement, whether it's engineering or not I can't say.

 

[Chestermiller]

That's my judgement, whether it's engineering or not I can't say.

I feel sorry for you.

 

[Chestermiller]

Does anyone want to complete this compression calculation with me so that we can show Pranj5 what the correct answer should be?
1. % quality vs pressure
2. change in enthalpy vs pressure (work vs pressure)

 

[Mech_Engineer]

Enthalpy is something more than just internal energy and it's very important factor in this whole process. Actually, I have started this thread to know whether steam will be in saturated state or not during compression with water and without considering, it just can't be solved.

For the purposes of this discussion you should consider Enthalpy a physical parameter of the fluid like temperature or pressure. I think you might be getting a little wrapped around the axle regarding its significance in a thermodynamic process. It is a combination of the internal energy of a working fluid, plus the product of pressure and volume of that fluid.

Chester has shown what I would consider a herculean amount of restraint and patience in this thread and I'm amazed it's stuck around this long actually. For your question regarding compression of a partly saturated fluid, I personally would stick to being a practical engineer and use a t-s , h-s. or possibly p-v diagram for steam. Take for example this one:

http://www.gemco.fr/medias/MollierDiag.png

Say I have 50% saturated steam at 10 Bar (approx 180 Celsius at this state) and want to know what would happen if I compressed to 100 Bar.

• As a first assumption, I would estimate the compression cycle to be isentropic and follow a constant-entropy line (vertical line on a T-s diagram) from state 1 (10 Bar) to state 2's pressure (100 Bar).
• A quick eye estimate looks like this would result in approximately 42% saturated steam at 100 Bar (with a resulting state 2 temperature of about 310 Celsius).
• This compression process also increases the specific enthalpy of the fluid from a little under 1800 kJ/kg to a little under 2000 kJ/kg, it looks to be a change of about 200 kJ/kg.

I personally prefer to see thermodynamic cycles in the context of these diagrams, so I can "visualize" what's going on; maybe it can help for you as well?

 

[pranj5]

Thank you Mr. Mech_Engineer. Kindly just tell me one thing. You have considered that during the whole process, the steam will remain saturated. And at that point, I agree with you. If the steam wouldn't be saturated and behave like an ideal gas, power consumption would be much higher than just 200 kJ/kg.
You just confirmed what I have assumed from the very beginning.

 

[Mech_Engineer]

If the steam wouldn't be saturated and behave like an ideal gas, power consumption would be much higher than just 200 kJ/kg.

I'm not sure what you mean by "wouldn't be saturated." Either you have superheated steam (which can be analyzed using the chart I referenced), a saturated mixture of steam and water (also can be analyzed using the above chart), or straight water. So which is it?

 

[pranj5]

I'm not sure what you mean by "wouldn't be saturated." Either you have superheated steam (which can be analyzed using the chart I referenced), a saturated mixture of steam and water (also can be analyzed using the above chart), or straight water. So which is it?

Sorry, as I was unable to express my thoughts properly. I mean if the steam become superheated. But, as long as there is water mixed with steam i.e. the steam is "wet", then it just simply can't be superheated and will remain saturated until all the water will be converted into steam. Am I right?
In short, I want to mean that compressing wet saturated steam would take much less power in comparison to dry, saturated steam.

 

[Mech_Engineer]

In short, I want to mean that compressing wet saturated steam would take much less power in comparison to dry, saturated steam.

Using the t-s diagram I provided, can you provide a simple analysis which tests your hypothesis? Maybe compare compressing 50% quality saturated steam vs saturated vapor and compare results?

 

[Mech_Engineer]

Since we're interested in Entropy vs. Enthalpy specifically, this diagram might make estimation easier:

http://docs.engineeringtoolbox.com/documents/308/mollier-diagram-water_2.png

Say we want to compress a fluid from 0.1 Bar to 1.0 Bar and are investigating the enthalpy change as our beginning state changes. Look what happens to the length of the violet arrows (which denote the enthalpy change) as we transition from starting as a saturated mixture to superheated vapor:

 

[Chestermiller]

Using the t-s diagram I provided, can you provide a simple analysis which tests your hypothesis? Maybe compare compressing 50% quality saturated steam vs saturated vapor and compare results?

Unfortunately, I don't think he knows how to use a t-s diagram to test his hypothesis.

 

[Mech_Engineer]

Unfortunately, I don't think he knows how to use a t-s diagram to test his hypothesis.

I thought of that after the fact, we posted at nearly the same instant trying to address this concern ;-)

 

[Chestermiller]

I thought of that after the fact, we posted at nearly the same instant trying to address this concern ;-)

You should have shown a case on your diagram where you start out with a combination of saturated liquid and vapor and end up with superheated vapor. Any chance you can add such a line (say starting out at 95% quality)? (Otherwise, we might think that you are proving his contention).

 

[Mech_Engineer]

No problem, I've added it as a beautiful fluorescent green arrow of justice!

 

[Mech_Engineer]

You should have shown a case on your diagram where you start out with a combination of saturated liquid and vapor and end up with superheated vapor. Any chance you can add such a line (say starting out at 95% quality)? (Otherwise, we might think that you are proving his contention).

I think it's clear based on the thermophysical charts provided and the example process arrows I've shown that reality is exactly opposite to @pranj5 's contention- the enthalpy difference increases when we are isentropically compressing superheated vapor compared to compressing a saturated mixture.

 

[pranj5]

I think it's clear based on the thermophysical charts provided and the example process arrows I've shown that reality is exactly opposite to @pranj5 's contention- the enthalpy difference increases when we are isentropically compressing superheated vapor compared to compressing a saturated mixture.

My contention is just what you have said in this post above.
Whatsoever, as per your diagram on post 65, it seems that if the steam is 80% steam and 20% water mixture, then the power consumption is 300 kJ/kg. When it's 90% steam and 10% water, then it's around 350 kJ/kg and so on. I hope I have correctly analysed your diagram.

 

[Mech_Engineer]

My contention is just what you have said in this post above.
Whatsoever, as per your diagram on post 65, it seems that if the steam is 80% steam and 20% water mixture, then the power consumption is 300 kJ/kg. When it's 90% steam and 10% water, then it's around 350 kJ/kg and so on. I hope I have correctly analysed your diagram.

That sounds about right, I thought for some reason you were making the argument that compressing a saturated mixture would require more energy than compressing superheated steam. In any case, I think this form of chart answers this question:

Actually, I have started this thread to know whether steam will be in saturated state or not during compression with water and without considering, it just can't be solved.

Charts of this kind are in my opinion the most reasonable method for modeling and rough numbers for most thermodynamic cycles. I personally find it beneficial to be able to "visualize" a thermodynamic process in the context of a t-s or P-v chart (although as we've seen an h-s chart can be very helpful as well).

 

[Chestermiller]

I think it's clear based on the thermophysical charts provided and the example process arrows I've shown that reality is exactly opposite to @pranj5 's contention- the enthalpy difference increases when we are isentropically compressing superheated vapor compared to compressing a saturated mixture.

Actually, his main contention has been that if you compress a saturated mixture of steam and liquid water adiabatically and reversibly, the quality of the mixture decreases(I.e., the mass fraction of liquid increases).

 

[Mech_Engineer]

Well I think its clear from the diagrams (and no doubt what Chester has been explaining the whole time) that isentropically compressing a [high quality] saturated mixture results in a higher quality mixture (a.k.a. a higher mass fraction of vapor to liquid).

 

[Chestermiller]

Also, regarding the other contention, it's obvious that, the closer the mixture is to being all liquid water, the less enthalpy change is going to be involved in increasing its pressure any given amount. Just think about how small the enthalpy change is in increasing the pressure on low specific volume pure liquid water.

 

[Mech_Engineer]

Well I think its clear from the diagrams (and no doubt what Chester has been explaining the whole time) that isentropically compressing a saturated mixture results in a higher quality mixture (a.k.a. a higher mass fraction of vapor to liquid).

It looks to me though that there is a transition, whether the quality increases or decreases depends on your starting saturated mixture. If you look at the following diagram again (which covers a larger quality range):

Starting at around x=60% quality, the isentropic compression behavior reverses (or near the middle starts aiming for the triple point rather than superheated vapor). With high mass fractions of vapor the increased pressure no doubt heats the net mixture such that the net mixture increases in quality; but in the low qualities (say x=20%) the increased pressure causes a net decrease in the quality. So depending on which half of the "bell-curve" you're starting in, increasing pressure isentropically will bring you towards the saturation curve; whether you end up with vapor or liquid at the end of the process step depends on which half of the curve you started in.

 

[Mech_Engineer]

Also, regarding the other contention, it's obvious that, the closer the mixture is to being all liquid water, the less enthalpy change is going to be involved in increasing its pressure any given amount. Just think about how small the enthalpy change is in increasing the pressure on low specific volume pure liquid water.

This can also be visualized in the above diagram, note that the blue constant-enthalpy curves approach vertical on the saturated liquid side of the curve. This means that an isentropic compression step (which follows a vertical line in this diagram) would result in nearly no enthalpy change since it would closely follow the constant-enthalpy lines.

 

[Chestermiller]

This can also be visualized in the above diagram, note that the blue constant-enthalpy curves approach vertical on the saturated liquid side of the curve. This means that an isentropic compression step (which follows a vertical line in this diagram) would result in nearly no enthalpy change since it would closely follow the constant-enthalpy lines.

Sure. For the pure liquid at constant entropy, dh is just vdP.

 

[Chestermiller]

Starting at around x=60% quality, the isentropic compression behavior reverses (or near the middle starts aiming for the triple point rather than superheated vapor). With high mass fractions of vapor the increased pressure no doubt heats the net mixture such that the net mixture increases in quality; but in the low qualities (say x=20%) the increased pressure causes a net decrease in the quality. So depending on which half of the "bell-curve" you're starting in, increasing pressure isentropically will bring you towards the saturation curve; whether you end up with vapor or liquid at the end of the process step depends on which half of the curve you started in.

Do you think the OP was really envisioning a large mass fraction of liquid >40% to begin with? If he was, I guess he should have mentioned it. My interpretation was that he was talking about an incremental change from saturated steam, involving something on the order of about 10% liquid. I'm wondering what a process and equipment starting the compression with a mixture having a large mass fraction of liquid would look like.

 

[pranj5]

That sounds about right, I thought for some reason you were making the argument that compressing a saturated mixture would require more energy than compressing superheated steam. In any case, I think this form of chart answers this question:
Charts of this kind are in my opinion the most reasonable method for modeling and rough numbers for most thermodynamic cycles. I personally find it beneficial to be able to "visualize" a thermodynamic process in the context of a t-s or P-v chart (although as we've seen an h-s chart can be very helpful as well).

In contrary, I want to say that as the steam will remain saturated, therefore much less energy will be needed to compress that in comparison to compressing superheated steam.

 

[Mech_Engineer]

It also means less energy is available to extract in a power generation application, paired with the fact that low quality steam is less useful in turbines subject to damage from the water droplets.

So, why do you care of it takes less energy to increase pressure of a saturated fluid mixture?

 

[pranj5]

It also means less energy is available to extract in a power generation application, paired with the fact that low quality steam is less useful in turbines subject to damage from the water droplets.
So, why do you care of it takes less energy to increase pressure of a saturated fluid mixture?

Simple. The quality of the high pressure water can be increased by heating it. Heat is the lowest form of energy and mechanical energy/electricity is the highest form. If, by using wet steam, we can reduce power consumption for steam compression, that means we are saving mechanical power/electricity at the expense of heat. Certainly that can be considered as profit.
And secondly, we can use this steam to heat up another fluid to drive a turbine or other kind of power generation machinery. In short, the steam produced can be used as heat source for another power generation system.
And, by the way, my initial thought was compressing steam inside a cylinder with some amount of water below. What I have thought that as the steam will always be in contact with water, it will always remain saturated (steam in contact with water will always remain saturated, right?) and therefore much less power will be needed to compress that.

 

[Mech_Engineer]

Please review the current state of technology regarding power generation cycles using steam as a working fluid before making broad assumptions about your "discovery" that heating water will turn it into vapor. https://en.wikipedia.org/wiki/Thermodynamic_cycle#Well-known_thermodynamic_cycles

Additionally, before making any claims about your idea regarding efficiency or energy use, you need to provide a diagram which shows it in a T-s diagram or similar as a complete cycle. I'm unable to do your analysis work for you.

https://en.wikipedia.org/wiki/Rankine_cycle

 

[pranj5]

Please review the current state of technology regarding power generation cycles using steam as a working fluid before making broad assumptions about your "discovery" that heating water will turn it into vapor. https://en.wikipedia.org/wiki/Thermodynamic_cycle#Well-known_thermodynamic_cycles

https://en.wikipedia.org/wiki/Thermodynamic_cycle#Well-known_thermodynamic_cycles
It's pretty simple. As for example, if wet steam has been compressed to 10 bars and it will become 20% wet steam at 180C. Now, what's the best way to convert the water present in the wet steam into steam. If the amount is 5 kg, that means it has 1 kg water inside and to convert that water into steam we have to add 2012.2 kJ of heat to the steam. Have you thought something else?

Additionally, before making any claims about your idea regarding efficiency or energy use, you need to provide a diagram which shows it in a T-s diagram or similar as a complete cycle. I'm unable to do your analysis work for you.

https://en.wikipedia.org/wiki/Rankine_cycle

It will be just like this:

This is the T-S diagram of open-cycle OTEC where seawater is evaporated and then used to run a turbine. In another version named closed cycle OTEC, the steam produced by this water is used to heat up a secondary fluid to rotate a turbine. I just want to mean that.

 

[Mech_Engineer]

I'm not quite remembering what the purpose of this thread is anymore, what specific input are you needing from us at this point? You seem to be proposing the use of a steam turbine for... what exactly? What's wrong with a typical Rankine-cycle process?

 

[pranj5]

I am not proposing use of steam for steam turbine but rather how to reduce power consumption during steam compression. At present, market available steam compressors use much more energy than and the steam will become highly superheated during the process. I just want to know whether having water with steam during the compression process can reduce power consumption by keeping the steam saturated during the process or not.
As you have stated already that wet, saturated steam takes much less energy to compress than superheated steam. I hope I have defined the purpose of this thread.

 

[Mech_Engineer]

I am not proposing use of steam for steam turbine but rather how to reduce power consumption during steam compression.

Steam compression/power consumption in the context of what thermodynamic cycle? For power generation purposes the Rankine Cycle does not compress steam, it compresses water with a pump and then heats it with a boiler.

There are four processes in the Rankine cycle. These states are identified by numbers (in brown) in the above T-s diagram.

• Process 1-2: The working fluid is pumped from low to high pressure. As the fluid is a liquid at this stage, the pump requires little input energy.
• Process 2-3: The high pressure liquid enters a boiler where it is heated at constant pressure by an external heat source to become a dry saturated vapour. The input energy required can be easily calculated graphically, using an enthalpy-entropy chart (aka h-s chart or Mollier diagram), or numerically, using steam tables.
• Process 3-4: The dry saturated vapour expands through a turbine, generating power. This decreases the temperature and pressure of the vapour, and some condensation may occur. The output in this process can be easily calculated using the chart or tables noted above.
• Process 4-1: The wet vapour then enters a condenser where it is condensed at a constant pressure to become a saturated liquid.

 

[pranj5]

Kindly leave thermodynamic cycles aside and just keep focusing on steam compression. Point is, how to reduce power consumption during steam compression process.

 

[Chestermiller]

Kindly leave thermodynamic cycles aside and just keep focusing on steam compression. Point is, how to reduce power consumption during steam compression process.

By this rationale, you can really minimize the power consumption by just taking pure liquid (no vapor) and increasing the pressure on it using a pump.

 

[pranj5]

What you are suggesting is liquid compression while what I want to do is steam compression. I know well that liquids can be compressed easily but that's not my motto. I just want to compress low pressure steam to higher pressure and want to keep the steam in saturated state during the process. What I know is if the steam will remain saturated during the process, compression will take much less power.

 

[Mech_Engineer]

What I know is if the steam will remain saturated during the process, compression will take much less power.

Take much less power for achieving what??

 

[pranj5]

For steam compression. If steam can be compressed with much less power, that means opening of new horizon in power generation.

 

[Chestermiller]

It's pretty simple. As for example, if wet steam has been compressed to 10 bars and it will become 20% wet steam at 180C.

This is the final state of the compressed wet steam. What do you envision as a typical initial pressure, temperature, and wetness of the steam for this situation?

 

[pranj5]

This is the final state of the compressed wet steam. What do you envision as a typical initial pressure, temperature, and wetness of the steam for this situation?

This has been written in reply to Mech_Engineer's post that wet steam has low quality and I just want to show him that the quality of steam at higher pressure can be increased by applying heat to it. I don't have any vision about the initial stage from where the steam has started. Just somehow its pressure has been increased and still it has water inside and therefore it's wet, saturated steam and can be considered as low quality.

 

[Chestermiller]

This has been written in reply to Mech_Engineer's post that wet steam has low quality and I just want to show him that the quality of steam at higher pressure can be increased by applying heat to it. I don't have any vision about the initial stage from where the steam has started. Just somehow its pressure has been increased and still it has water inside and therefore it's wet, saturated steam and can be considered as low quality.

I thought all along you have been talking about compressing it adiabatically, without adding any heat.

 

[pranj5]

Kindly note that heat will be added here after the compression being completed. Not before or during the process.

 

[Mech_Engineer]

For steam compression. If steam can be compressed with much less power, that means opening of new horizon in power generation.

I'm sorry to say this is incorrect. Thermodynamic cycles for power generation do not depend on compressing steam, as I've already pointed out in the discussion of the Rankine cycle which is used in power stations worldwide. The efficiency of the Rankine cycle is driven by the amount of work the turbine can extract, which is defined as step 3-4 in the diagram:

Typical methods for increasing the efficiency of this cycle are superheating the steam for a higher temperature starting point (higher carnot efficiency), and regenerative cycles which take advantage of residual heat in the tubine's exhaust with multi-stage expansion systems (in conjunction with superheating).

 

[pranj5]

I'm sorry to say this is incorrect.

If steam can be compressed to higher pressure from lower pressure, that will have many many applications than just power generation. Steam is used in many kind of industries and a huge amount is wasted at present as steam compressors will consume more energy and they are costly and therefore making new steam by using conventional fuel is cheaper in comparison to compressing low pressure steam to higher pressure. But, if the power consumption can be reduced by any means, much less fuel will be burnt and that means lower GHG emission.
As for example, at present it will take around 800 kW (multiple stage compressor with intercooler) to compress steam from 2.536 kPa (saturated steam pressure at 20C) to 1 bar @ 1 kg/sec. While the amount of heat embedded in 1 kg/sec steam flow is around 2.6 MW. Now, if we consider that the electricity comes from a thermal power plant with around 33% efficiency, that means it will take 2.4 MW of heat to produce 800 kW of electricity and at the end we will just have left with 200 kW of heat, a mere return against huge investment in steam compressor. But, if the steam will remain in saturated state during the whole process, that amount can be reduced to just 200 kW and you yourself can calculate the increase in the amount of additional heat which can make the investment in steam compressors more attractive and can save huge amount of fuel.

 

[Mech_Engineer]

But, if the steam will remain in saturated state during the whole process, that amount can be reduced to just 200 kW and you yourself can calculate the increase in the amount of additional heat which can make the investment in steam compressors more attractive and can save huge amount of fuel.

I think its time you calculate the alleged savings before making any more claims. Use a steam chart of your choosing, maybe the h-s chart we were looking at before.

 

[Chestermiller]

If steam can be compressed to higher pressure from lower pressure, that will have many many applications than just power generation. Steam is used in many kind of industries and a huge amount is wasted at present as steam compressors will consume more energy and they are costly and therefore making new steam by using conventional fuel is cheaper in comparison to compressing low pressure steam to higher pressure. But, if the power consumption can be reduced by any means, much less fuel will be burnt and that means lower GHG emission.
As for example, at present it will take around 800 kW (multiple stage compressor with intercooler) to compress steam from 2.536 kPa (saturated steam pressure at 20C) to 1 bar @ 1 kg/sec. While the amount of heat embedded in 1 kg/sec steam flow is around 2.6 MW. Now, if we consider that the electricity comes from a thermal power plant with around 33% efficiency, that means it will take 2.4 MW of heat to produce 800 kW of electricity and at the end we will just have left with 200 kW of heat, a mere return against huge investment in steam compressor. But, if the steam will remain in saturated state during the whole process, that amount can be reduced to just 200 kW and you yourself can calculate the increase in the amount of additional heat which can make the investment in steam compressors more attractive and can save huge amount of fuel.

You seem so very confident about all this. And, since you are so confident, what was your reason for starting this thread in the first place?