Superheated steam and water together

[Chestermiller]

f(20) = 0.01683 - 0.02740 = -0.0106
f(28) = 0.02650 - 0.01310 = +0.01340
f(24) = 0.02119 - 0.02027 = 0.00092
From plotting of these results on a graph, I have found that the solution for the final temperature and the final mass fraction of steam are:

T = 23.6 C
x = 0.0207

So, the final mass of liquid water is $m_L= 1.006 kg$, and the final mass of water vapor is $m_V=0.02127 kg$. So, of the 10 grams of superheated steam that is added, 6 grams end up in the liquid and 4 grams end up in the vapor.

 

[pranj5]

Well, thank you for your great effort. Whatsoever, at the end we can conclude that a part of the injected steam will be added to the water inside to raise its temperature and both pressure and temperature inside will rise and will come to equilibrium in a saturated state.
It's a lengthy process and there is no formula available to calculate the final temperature and pressure. I am curious about another fact. If, instead of injecting steam, the vapour inside will be compressed in such a way that the density of steam will increase from 20C level to 23.6C level, the same thing will occur. In fact, injecting a little amount of steam inside means compressing the already available steam inside a little, right?

 

[Chestermiller]

Well, thank you for your great effort. Whatsoever, at the end we can conclude that a part of the injected steam will be added to the water inside to raise its temperature and both pressure and temperature inside will rise and will come to equilibrium in a saturated state.
It's a lengthy process and there is no formula available to calculate the final temperature and pressure. I am curious about another fact. If, instead of injecting steam, the vapour inside will be compressed in such a way that the density of steam will increase from 20C level to 23.6C level, the same thing will occur. In fact, injecting a little amount of steam inside means compressing the already available steam inside a little, right?

We can analyze this quantitatively using the steam tables. In particular, we can elucidate the separate contributions of the original tank contents and injected superheated steam to the final overall results.

Imagine that there is a thin, massless, frictionless, insulated piston separating the original contents of the tank from the injected steam. Initially, the original contents take up the entire tank, but as time progresses (and we introduce the superheated steam very slowly into the space above the piston), the superheated steam compresses the original contents, while itself expanding against the piston. In the end, we reach the same final thermodynamic state for the system (when the original contents and the injected steam are both at 23.6 C and the corresponding equilibrium vapor pressure, which we can determine from the steam tables). We then remove the piston barrier from between the two regions and allow whatever liquid water is formed in the overhead space to merge with the liquid water below.

What this analysis will show is that (before the barrier is removed) there will be a change in the amount of liquid water both within the original contents and within the contents of the overhead space. The sum of these changes (i.e., the net change) will be add up to the 6 grams we determined previously. However, neither of these changes will be zero. There will also be corresponding changes in the mass of water vapor in each of the two regions which will add up to the 4 grams we determined previously.

Are you game to carry out this more detailed analysis?

Chet

 

[pranj5]

Well, at least I can say that what the superheated steam will do can be done mechanically and the result will be the same, right?

 

[Chestermiller]

Well, at least I can say that what the superheated steam will do can be done mechanically and the result will be the same, right?

No. What do you think will happen to the amount of liquid water in the container if we adiabatically compress the original contents with a piston to the equilibrium vapor pressure of water at 23.6 C?

 

[pranj5]

I just forgot to add the word adiabatic.

 

[Chestermiller]

I just forgot to add the word adiabatic.

So, what do you think will happen to the amount of liquid water in the container if we compress the original contents of the container with a piston to the equilibrium vapor pressure of water at 23.6 C?
(a) it will increase
(b) it will remain the same
(c) it will decrease

Choose a, b, or c.

 

[pranj5]

If you want to mean that if by mechanical means, the pressure has been increased from 20C saturation level to 23.6C saturation level; then the answer is very much evident. Kindly read all the posts (especially posts by you) to any third person and ask him/her the same question and let's see the answer.

 

[Chestermiller]

If you want to mean that if by mechanical means, the pressure has been increased from 20C saturation level to 23.6C saturation level; then the answer is very much evident. Kindly read all the posts (especially posts by you) to any third person and ask him/her the same question and let's see the answer.

Well, compressing it mechanically with a piston is the same as compressing it using the superheated steam (as you yourself pointed out), provided the original contents is kept separate from the injectate using a massless, frictionless, insulated piston to separate them.

So, please, refresh my memory: choose (a), (b), or (c).

I will help you check your answer using the steam tables to quantify it precisely.

 

[Chestermiller]

I just wanted to see what you were going to say. I think you were going to say that the amount of liquid water increases. Surprise! I tend to agree with this assessment for this particular (very wet) case in which the amount of liquid water is much, much, higher than the amount of water vapor. But we would like to quantify precisely how much the increase would be by employing the steam tables. The calculation is really simple.

You said in a previous thread that, for the adiabatic reversible compression of a steam/water mixture, the change in entropy of the mixture should be zero. So let's start out by calculating the entropy of the original mixture. At 20 C, the steam tables give values of the specific entropy of saturated liquid water and saturated water vapor as $s_L=0.2966 kJ/kgK$ and $s_V=8.667 kJ/kgK$, respectively. Therefore, the total initial entropy of our system is given by:
$$S=(1)(0.2966)+(0.0173)(8.667)=0.4198\ kJ/K$$
This must also be the total entropy of our final saturated mixture at 23.6 C. Interpolating in the steam tables to a saturation temperature of 23.6 C, we calculate values of the specific entropy of saturated liquid water and saturated water vapor as $s_L(23.6)=0.3477\ kJ/kgK$ and $s_V(23.6)=8.588\ kJ/kgK$. Therefore, for the mixture at 23.6 C, we have$$1.0173[0.3477+(8.855-0.3477)x]=0.4198\ kJ/K$$where x is the final mass fraction of water vapor. Solving for x yields x = 0.00764. So, the final mass of saturated liquid water at 23.6 C is 1.0095 gm. Thus, the mass of liquid water increases by 9.5 grams; and the mass of water vapor decreases by 9.5 grams to 7.8 grams. This compares with the 6 gm increase that is obtained when we inject 10 gm of superheated steam at 60 C into the tank.

 

[pranj5]

You said in a previous thread that, for the adiabatic reversible compression of a steam/water mixture, the change in entropy of the mixture should be zero.

I haven't said that. I just want to consider this for the sake of simplicity. And can you tell me how the amount of water can be a factor here. I mean if the amount of water is sufficiently high and more than the amount of steam, then amount of water will increase and will decrease if the case is reverse.
One simple explanation is that, if case of lesser amount of water, additional heat generated by the inefficiency of the process will evaporate it. That can certainly happen in reality. But, at least we both are agreed now on one point that if the amount of water is sufficient, then the steam will remain saturated during the process of compression. Even if the compression is adiabatic. Right?

 

[Chestermiller]

I haven't said that. I just want to consider this for the sake of simplicity. And can you tell me how the amount of water can be a factor here. I mean if the amount of water is sufficiently high and more than the amount of steam, then amount of water will increase and will decrease if the case is reverse.
One simple explanation is that, if case of lesser amount of water, additional heat generated by the inefficiency of the process will evaporate it. That can certainly happen in reality. But, at least we both are agreed now on one point that if the amount of water is sufficient, then the steam will remain saturated during the process of compression. Even if the compression is adiabatic. Right?

Yes, as long as the initial wetness is greater than about 50%.

 

[pranj5]

Do you mean that if the amount of water is more than the amount of steam, then the steam will remain saturated during the compression process? I hope this is what you want to mean by initial wetness, right?

 

[Chestermiller]

Do you mean that if the amount of water is more than the amount of steam, then the steam will remain saturated during the compression process? I hope this is what you want to mean by initial wetness, right?

Yes. Also, even if it's not more, it will still stay saturated if you limit the amount of compression.

 

[pranj5]

In that case, what should be the limit of compression ratio?

 

[Chestermiller]

In that case, what should be the limit of compression ratio?

It depends of the fraction of liquid water. To find that out, it is easiest to use the temperature-entropy diagram or the enthalpy-pressure diagram.

Incidentally,
You're welcome.

 

[pranj5]

Can you guess some limit?

 

[Chestermiller]

Can you guess some limit?

There are a lot of factors involved. Can you provide a specific initial state?

 

[pranj5]

Lets start with our old initial states i.e. saturated steam at 20C and 2.536 kPa.

 

[Chestermiller]

Lets start with our old initial states i.e. saturated steam at 20C and 2.536 kPa.

I'll check the Mech_Engineer's T-S diagram, and get back with you.

 

[pranj5]

How the compression limit can be determined by the T-S diagram?

 

[Chestermiller]

How the compression limit can be determined by the T-S diagram?

You move up a constant entropy line until you hit the 100% dry envelope. You determine the temperature at that point. Then you determine the corresponding saturation vapor pressure, and take the ratio. For a temperature higher than that, the compressed gas would be superheated (i.e., beyond your desired limit).

 

[pranj5]

If the amount of water increases with pressure, then how can we hit the 100% dry envelope?

 

[Chestermiller]

If the amount of water increases with pressure, then how can we hit the 100% dry envelope?

The amount of liquid water increases with pressure if there isn't enough liquid water around to start with. Mech_Engineer showed this (i.e., if it's pretty wet to start with). Otherwise, the mixture will become drier as you compress. So the amount of water increasing or decreasing depends on the initial wetness. Check out the T-S diagram.

 

[pranj5]

The amount of liquid water increases with pressure if there isn't enough liquid water around to start with.

That's contrary to what you have said in post 42.

 

[Chestermiller]

That's contrary to what you have said in post 42.

I'm going to try to say this as clearly as I can.

1. If the initial mixture is low enough quality (high wetness), the amount of liquid increases as you compress the mixture no matter how much you compress.
2. If the initial mixture is high quality (low wetness), the amount of liquid decreases as you compress the mixture, and the mixture can reach 100% dryness (if you compress it enough). But, if you don't compress it the necessary amount, you will still have some liquid left over in the end.
3. There is a cross over in these two kinds of behavior at an initial quality in the 40 - 60 % range (depending on the starting temperature/pressure).

I strongly urge you to spend time examining the temperature entropy diagram for water so that you can see for yourself how this all works out.

 

[pranj5]

Thanks! Though the actual scenario isn't much clear. Whatsoever, at least it can be said that with sufficient water, the steam will remain saturated during the compression process and part of it will be added to the water to increase the enthalpy of water to reach to saturation level of higher temperature. In such case, power needed to compress that steam will be less as it will remain saturated and power needed to compress steam will be least if the steam can be preserved at saturated state during the process.

 

[pranj5]

Well, the amount of water can be calculated if we consider that the amount of steam will be mixed with water sufficient amount of water so that all the latent heat of steam will add to the final volume and raise its temperature and pressure. As for example, if we take 1 kg steam at 20C and have to mix it with water by pressurisation so that the latent heat of the steam will be contained in the final volume of water, then we will need 32 kg of water for that.

 

[pranj5]

Let's discuss this matter from another point of view. What will happen if instead of compressing the steam, the water will be heated (consider the whole system to be totally insulated). First small amount of steam will rise from the water and this will add to the amount of steam inside and thus increasing the pressure and temperature inside and at the same time the boiling point of the water will increase too. In short, the boiling point of the water increases along with the pressure and temperature inside and at the same time both water and steam will remain in saturated state.
In fact, what I have said above is the principle behind heat pipe, which are well tested and now is even used in space too.

 

[Nidum]

Boiler ? As in what supplies steam to a steam engine ?