Compressing steam and water together

[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.

 

[pranj5]

Just to clarify myself, I want to answer your questions:
Step !: C(nL+nV)(TR-T)
Step 2: If the process is adiabatic, then in fact there will be no change in enthalpy. The gross enthalpy of pressurised water will be divided into enthalpy of leftover water and enthalpy of the steam. There will be change in enthalpy if some kind of work has to be done by pushing the piston. Assuming the piston to be weightless and frictionless,
Step 3: λ(T)nv is the change in enthalpy of the amount of water that has been changed into steam.
Whatsoever, just want to remind you that you are considering λ(T) to be the latent heat of vaporisation of gm-mole of water, while during the calculation you have put the value of latent heat to be per gm.

 

[Chestermiller]

Just to clarify myself, I want to answer your questions:

You are aware that enthalpy is a unique physical property of a material (i.e., a function of state), in this case water, independent of any particular process imposed on the material, correct? What we are doing here is that, independent of our actual adiabatic reversible process involving liquid water and steam, we are basically conducting a separate (thought) experiment to quantify the effect of temperature, pressure, and mass fraction steam on the enthalpy of a mixture of liquid water and steam. This thought experiment does not have to bear any resemblance whatsoever to the actual process we are analyzing.

Step !: C(nL+nV)(TR-T)

No. It should be $C(n_L+n_V)(T-T_R)$. $T_R$ corresponds to the initial state and T corresponds to the final state.

Step 2: If the process is adiabatic, then in fact there will be no change in enthalpy. The gross enthalpy of pressurised water will be divided into enthalpy of leftover water and enthalpy of the steam. There will be change in enthalpy if some kind of work has to be done by pushing the piston. Assuming the piston to be weightless and frictionless,

No. The change in enthalpy per mole of an incompressible liquid is given by $C\Delta T +v\Delta P$. In the case of Step 2, $\Delta T =0$ and $\Delta P=(P-P_R)$. So, in Step 2, the change in enthalpy of our liquid is $(n_L+n_V)v_L(P-P_R)$

Step 3: λ(T)nv is the change in enthalpy of the amount of water that has been changed into steam.

Correct.

Whatsoever, just want to remind you that you are considering λ(T) to be the latent heat of vaporisation of gm-mole of water, while during the calculation you have put the value of latent heat to be per gm.

It doesn't matter which you use as long as you do it in a mathematically consistent way. In the derivation of the equations, I've been using moles, but in the calculations, I'm equally comfortable using kg. If this causes you discomfort, we can do everything in one or the other. Which do you prefer?

Taking into account my corrections to your answers, what is the equation for the enthalpy H of a mixture of liquid water and steam in saturated State 2 relative to the reference state R?

 

[pranj5]

No. It should be $C(n_L+n_V)(T-T_R)$. $T_R$ corresponds to the initial state and T corresponds to the final state.

I myself too want to say that. I just misunderstanding of representation.

No. The change in enthalpy per mole of an incompressible liquid is given by $C\Delta T +v\Delta P$. In the case of Step 2, $\Delta T =0$ and $\Delta P=(P-P_R)$. So, in Step 2, the change in enthalpy of our liquid is $(n_L+n_V)v_L(P-P_R)$

In this case, ΔT isn't zero. The vapour produced go the latent heat from the water and its temperature lowered. And it's now converted into steam and water and therefore can't be considered to be something totally in-compressible now. Water part can be considered non-compressible, but not the steam.

It doesn't matter which you use as long as you do it in a mathematically consistent way. In the derivation of the equations, I've been using moles, but in the calculations, I'm equally comfortable using kg. If this causes you discomfort, we can do everything in one or the other. Which do you prefer?

How can you be mathematically consistent, when you consider a value to be calorie/gm-mole and during calculation you have put it to be calorie/gm.

Taking into account my corrections to your answers, what is the equation for the enthalpy H of a mixture of liquid water and steam in saturated State 2 relative to the reference state R?

The gross enthalpy will be the same, what will happen is that the gross enthalpy will then be divided into two part; gross enthalpy of water and gross enthalpy of steam part. To change the enthalpy, either the steam has to perform some work (loss in enthalpy) and have to add heat/energy (gain in enthalpy) to the system. Otherwise, it will be same if the process is frictionless and weightless.

 

[Chestermiller]

In this case, ΔT isn't zero. The vapour produced go the latent heat from the water and its temperature lowered. And it's now converted into steam and water and therefore can't be considered to be something totally in-compressible now. Water part can be considered non-compressible, but not the steam.

As I said. This is a separate thought experiment that we are doing, not the actual process. There is no steam formed in Step 2. The pressure in this step is always greater than the equilibrium vapor pressure (so we have liquid water present throughout). Only at the very end of this step do we reach the equilibrium vapor pressure, but at that point Step 2 ends. So, at the end of Step 2, we still have all liquid water, with the potential to start forming water vapor if we add some heat or drop the pressure a little more.

How can you be mathematically consistent, when you consider a value to be calorie/gm-mole and during calculation you have put it to be calorie/gm.

Like I said, if you are uncomfortable with this (for whatever reason), we can do everything in whatever set of units you prefer. Just pick the units you want, and we'll do the problem in those units (including the calculations).

The gross enthalpy will be the same, what will happen is that the gross enthalpy will then be divided into two part; gross enthalpy of water and gross enthalpy of steam part. To change the enthalpy, either the steam has to perform some work (loss in enthalpy) and have to add heat/energy (gain in enthalpy) to the system. Otherwise, it will be same if the process is frictionless and weightless.

I have no idea what you are saying here, but you seem to be uncomfortable with how you think I will continue the analysis.

However, so far you have accepted Eqn. 2, and, when you accept what I have said about Eqn. 3, we can continue the analysis. Please understand that Eqns. 2, 3, and 4 are designed to automatically capture everything you have been worrying about in the above paragraph. After you have accepted these three equations, the rest of the analysis is going to be straight mathematics. And we will have to let the chips fall where they may. So speak up now if you have further discomfort with these three equations.

 

[pranj5]

As I said. This is a separate thought experiment that we are doing, not the actual process. There is no steam formed in Step 2. The pressure in this step is always greater than the equilibrium vapor pressure (so we have liquid water present throughout). Only at the very end of this step do we reach the equilibrium vapor pressure, but at that point Step 2 ends. So, at the end of Step 2, we still have all liquid water, with the potential to start forming water vapor if we add some heat or drop the pressure a little more.

Vapour will start to form at the very moment when there will be some space available above water. If you consider water to be non-compressible, then space will always be formed over water when the pressure will be below atmospheric inside the cylinder.

Like I said, if you are uncomfortable with this (for whatever reason), we can do everything in whatever set of units you prefer. Just pick the units you want, and we'll do the problem in those units (including the calculations).

It's not a matter of whether I am uncomfortable or not, but rather consistency during calculations.

I have no idea what you are saying here, but you seem to be uncomfortable with how you think I will continue the analysis.

What I want to say is clear. Enthalpy means gross energy content of a system. If no external energy enters the system or the system have to perform some work, gross enthalpy will remain the same. That's basic physics.

However, so far you have accepted Eqn. 2, and, when you accept what I have said about Eqn. 3, we can continue the analysis. Please understand that Eqns. 2, 3, and 4 are designed to automatically capture everything you have been worrying about in the above paragraph. After you have accepted these three equations, the rest of the analysis is going to be straight mathematics. And we will have to let the chips fall where they may. So speak up now if you have further discomfort with these three equations.

Problem with mathematics is that, if you forgot the reality, then it will carry you straight to wrong conclusion. 2+3 = 5, but if you concluded 2 ships and 3 cows equals to 5 people, that's dangerous. Eqn 1,2 is very basic physics, but I have doubt about 3.
Whatsoever, our main problem that we are discussing here is whether steam will remain saturated or not if compressed with water. But, such equations have taken us far away from this. In one of your previous posts, you have said that steam will liquefy during expansion and that's contrary to very basic physics.

 

[Chestermiller]

Vapour will start to form at the very moment when there will be some space available above water. If you consider water to be non-compressible, then space will always be formed over water when the pressure will be below atmospheric inside the cylinder.

No way. What ever gave you the strange idea that atmospheric pressure matters in a system containing only water? Without adding heat, vapor will not form until the pressure drops to slightly below the equilibrium vapor pressure. As long at the pressure is at or above the equilibrium vapor pressure, vapor will not form.

It's not a matter of whether I am uncomfortable or not, but rather consistency during calculations.

Stop whining and choose a set of units to use.

What I want to say is clear. Enthalpy means gross energy content of a system. If no external energy enters the system or the system have to perform some work, gross enthalpy will remain the same. That's basic physics.

In our actual problem, the surroundings are doing work on the system to compress it. That is captured in Eqns. 1 and 2.

The gross energy content of a closed system is the internal energy U, not the enthalpy H. The enthalpy is equal to the internal energy U plus PV. If there is no work or heat exchanged with the surroundings, U cannot change. But, even without exchanging significant heat or work with the surroundings, the enthalpy of an incompressible liquid can change if the pressure changes, since, even if V is constant, $\Delta PV$ is not.

Whatsoever, our main problem that we are discussing here is whether steam will remain saturated or not if compressed with water. But, such equations have taken us far away from this. In one of your previous posts, you have said that steam will liquefy during expansion and that's contrary to very basic physics.

That's contrary to your concept of the basic physics, not mine. Whatsoever, are you willing to accept the results of the analysis once we have agreed upon Eqns. 2-4, which capture all the basic physics in an unbiased way.

 

[pranj5]

No way. What ever gave you the strange idea that atmospheric pressure matters in a system containing only water? Without adding heat, vapor will not form until the pressure drops to slightly below the equilibrium vapor pressure. As long at the pressure is at or above the equilibrium vapor pressure, vapor will not form.

Vapour will form, as I have already said before, whenever there will be empty space between piston and the water surface. At atmospheric pressure level, external and internal pressure will be same and below that space will form over water surface. That will be the case whatever may be the atmospheric pressure. It's simple basic physics and common sense.

Stop whining and choose a set of units to use.

Whatever may be the unit, those aren't same. Latent heat of vaporisation of water is 540 cal/gm and (540X18) cal/gm-mole i.e. 9720 cal/gm-mole.

In our actual problem, the surroundings are doing work on the system to compress it. That is captured in Eqns. 1 and 2.

Not the surroundings, but rather external mechanical force. Whatsoever, we both have agreed on 1 and 2, so it doesn't matter.

The gross energy content of a closed system is the internal energy U, not the enthalpy H. The enthalpy is equal to the internal energy U plus PV. If there is no work or heat exchanged with the surroundings, U cannot change. But, even without exchanging significant heat or work with the surroundings, the enthalpy of an incompressible liquid can change if the pressure changes, since, even if V is constant, $\Delta PV$ is not.

The gross energy content of a system isn't just the internal energy, but also the "the amount of energy required to make room for it by displacing its environment and establishing its volume and pressure" (https://en.wikipedia.org/wiki/Enthalpy) and that's enthalpy.

That's contrary to your concept of the basic physics, not mine. Whatsoever, are you willing to accept the results of the analysis once we have agreed upon Eqns. 2-4, which capture all the basic physics in an unbiased way.

None of your calculations so far has been able to show that steam will liquefy during expansion here. While even a school student can say that the opposite is correct. The question isn't here "personal" basic physics but rather what we can conclude from experiments. In normal life, we have seen that water will evaporate when pressure is reduced over it. In school book experiments, it has been clearly stated that temperature of water will drop and eventually it will freeze to ice when pressure will be reduced over it gradually.

 

[Chestermiller]

Vapour will form, as I have already said before, whenever there will be empty space between piston and the water surface. At atmospheric pressure level, external and internal pressure will be same and below that space will form over water surface. That will be the case whatever may be the atmospheric pressure. It's simple basic physics and common sense.

Whatever may be the unit, those aren't same. Latent heat of vaporisation of water is 540 cal/gm and (540X18) cal/gm-mole i.e. 9720 cal/gm-mole.

Not the surroundings, but rather external mechanical force. Whatsoever, we both have agreed on 1 and 2, so it doesn't matter.

The gross energy content of a system isn't just the internal energy, but also the "the amount of energy required to make room for it by displacing its environment and establishing its volume and pressure" (https://en.wikipedia.org/wiki/Enthalpy) and that's enthalpy.

None of your calculations so far has been able to show that steam will liquefy during expansion here. While even a school student can say that the opposite is correct. The question isn't here "personal" basic physics but rather what we can conclude from experiments. In normal life, we have seen that water will evaporate when pressure is reduced over it. In school book experiments, it has been clearly stated that temperature of water will drop and eventually it will freeze to ice when pressure will be reduced over it gradually.

As a Physics Forums Mentor and one of Physics Forums leading experts in Thermodynamics, I am shocked by your bogus responses. They display a total lack of understanding and ignorance of even the most fundamental concepts in Thermodynamics. Ideas that you have referred to as "common sense and basic physics" are actually totally incorrect. I am done wasting my time trying to deal with you. You're on your own now. This thread is hereby closed.

 

[Chestermiller]

I have reopened this thread, based on recent collaboration between Pranj5 and myself. When the thread closed, we were working on a discussion of the analysis I present in post #30. But there was disagreement over the equation for the enthalpy of a saturated mixture of water vapor and liquid water at saturation temperature T and corresponding saturation pressure P, relative to the enthalpy of the same amount of liquid water at $T=T_R=0 C$ and $P=P_R=1 bar$. This was Eqn. 3 in post #30. Before the thread was closed, we were able to reach consensus on the first and third terms in the equation, but not the middle term involving the pressure. I had indicated in post #36 that the middle term could be obtained by evaluating the change in enthalpy between the following two thermodynamic equilibrium states:

State A: $(n_L+n_V)$ moles of liquid water at temperature T and 1 bar

State B: $(n_L+n_V)$ moles of liquid water at temperature T and the saturation vapor pressure P at temperature T

Pranj5 doubted the possibility that thermodynamic equilibrium State 2 could even exist. However, now, as a result of our collaboration in PMs, he has reached the following determination: "If the water inside the cylinder will be at 20C, then no empty space can be formed over the water until the pressure inside will reach 2.536 kPa i.e. the saturated steam pressure of water at 20C." This is just a special case of State B for T = 20 C. So his conclusion also applies to other saturation temperatures and pressures as well.

Now, to determine the middle (pressure dependent term) in Eqn. 3, we need to evaluate the change in enthalpy between thermodynamic equilibrium State A and thermodynamic equilibrium State B above. To accomplish this, we make use of the following general enthaly equation for a single phase solid, liquid, or gas (which appears in virtually every Thermodynamics book):
$$\left(\frac{\partial h}{\partial P}\right)_T=v\left[1-T\left(\frac{\partial \ln{v}}{\partial T}\right)_P\right]\tag{3a}$$
where h is the enthalpy per mole and v is the molar specific volume. If we make the reasonable approximation that liquid water is nearly incompressible (i.e., v=const with respect to changes in temperature and pressure) over the range of pressures and temperature of interest in our compression process, Eqn. 3a can be integrated immediately at constant temperature from State A to State B to yield: $$H_B-H_A=(n_L+n_V)(h_B-h_A)=(n_L+n_V)v(P-P_R)$$
Note that, for the equilibrium saturation vapor pressure P less than $P_R=1 bar$, this term is negative. Note also that the overall contribution of this term is typically very small compared to the other terms in our analysis, and it is often neglected. Finally, its effect cancels completely with another term in our subsequent analysis.

This completes the development of Eqn. 3 in post #30.

This is where I'll stop for now.

 

[pranj5]

Now, question is, if the term is negative, where the extra enthalpy has gone?

 

[Chestermiller]

Now, question is, if the term is negative, where the extra enthalpy has gone?

You are aware that, from the definition of enthalpy, $\Delta h=\Delta u+\Delta (Pv)$, correct?

 

[pranj5]

I know that. But, question is, where this lost enthalpy has gone. As per the first law of thermodynamics, it should be added to somewhere else, right?

 

[Chestermiller]

I know that. But, question is, where this lost enthalpy has gone. As per the first law of thermodynamics, it should be added to somewhere else, right?

Please be patient. We'll see what the first law of thermodynamics tells us shortly.

Now, if, in the change from State A to State B, the change in internal energy $\Delta u$ is equal to zero, what is $\Delta h$ equal to?

 

[pranj5]

Δh

here is a negative quantity. That means as the pressure is reduced and the volume remain unchanged, PV has been reduced.

 

[Chestermiller]

here is a negative quantity. That means as the pressure is reduced and the volume remain unchanged, PV has been reduced.

Excellent. So, do you still want to go back and look at the first law of thermodynamics, or are you now satisfied with the result in post #44?

 

[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?