Superheated steam and water together

[pranj5]

Lets start with an imaginary experiment. There is an enclosed container having a wide (i.e. having sufficient surface area) bowl filled with water at the bottom. The container is in vacuum and filled only with saturated steam at that specific temperature. Both water and steam are at the same temperature and pressure.
Now, some amount of superheated steam has been injected into that container. What will happen then?
As per my guess, a part of the hot superheated steam will enter the water and will raise its temperature and soon the pressure and temperature inside will rise and the superheated stem will start evaporating water from the bowl until the condition inside will come to an equilibrium and all the steam will become saturated.
By this process, superheated steam will loose some heat that will be used to evaporate some water from the bowl to make all the steam insdie saturated.
Am I right?

 

[BvU]

Yes. Within certain limits. In equilibrium the vapour is saturated.
With a simple enthalpy balance you can calculate the final p and T.

 

[pranj5]

Yes. Within certain limits. In equilibrium the vapour is saturated.
With a simple enthalpy balance you can calculate the final p and T.

Thanks! But, I am curious to know how long it will take and what kind of factors can affect this time span.

 

[BvU]

Can't say much about the speed. @Chestermiller perhaps ?

 

[pranj5]

OK. Let's discuss it and try to find out the time necessary.
Actually, I want to say that both steam will condense and water will evaporate in the process. That may look contradictory but it will be fact in such a case. High energy steam molecules from the superheated steam will enter the water and will raise its temperature. At the same time, comparative low energy steam molecules will rise and will mix up with supherheated steam to drag it down to saturated state. Am I right?

 

[Chestermiller]

OK. Let's discuss it and try to find out the time necessary.
Actually, I want to say that both steam will condense and water will evaporate in the process. That may look contradictory but it will be fact in such a case. High energy steam molecules from the superheated steam will enter the water and will raise its temperature. At the same time, comparative low energy steam molecules will rise and will mix up with supherheated steam to drag it down to saturated state. Am I right?

Why speculate? Why don't you just do the calculation (and thereby remove all the uncertainty)?

 

[pranj5]

Why speculate? Why don't you just do the calculation (and thereby remove all the uncertainty)?

It's not speculation, but rather trying to envisioning the whole process first and then go for the next step. And, by the way, can you suggest how to calculate the time necessary for evaporation?

 

[Chestermiller]

It's not speculation, but rather trying to envisioning the whole process first and then go for the next step. And, by the way, can you suggest how to calculate the time necessary for evaporation?

Before calculating the time required for the equilibration, I strongly recommend that you first determine (quantitatively) the final equilibrium state. Why? If you are unable to model that simpler situation, you will never be able to analyze the much more complicated problem of the time-dependent behavior (in my judgment). Plus, the solution for the final equilibrium state might suggest some ideas on how the system will behave quantitatively in the transient situation. Are you prepared to work with me on solving for the final equilibrium state?

 

[pranj5]

No problem!

 

[Chestermiller]

No problem!

OK. To get us started, I need you to specify a focus problem for us to analyze:

Initial Conditions in Tank
1. Volume of tank
2. Temperature in tank
3. Mass fraction of liquid water in tank

Conditions of Superheated Injectate
1. Temperature
2. Pressure
3. Mass of superheated steam injected into tank

 

[Ali Durrani]

this whole process that you have said what is the surety level that this will happen exactly as you said ? it depends on the amount of water in the container the mass of superheated steam the degree of super heat, and mass of liquid etc, you can't say this before even solving for the final conditions

 

[Mech_Engineer]

Now, some amount of superheated steam has been injected into that container. What will happen then?

Are you planning to model this step as a constant-pressure process? Adding mass will have to increase pressure as well for a fixed-volume container.

What you're describing sounds a little like a flash evaporation process in a vapor-liquid separator.

 

[pranj5]

this whole process that you have said what is the surety level that this will happen exactly as you said ? it depends on the amount of water in the container the mass of superheated steam the degree of super heat, and mass of liquid etc, you can't say this before even solving for the final conditions

As per my guess, more steam will be added to the water than the amount of water evaporated. Because, only in that case the temperature of water can rise. You are right that it will certainly be dependent on the degree of superheating and also the amount of superheated steam being injected. If the temperature is too high (above triple point) and the amount too is high and more than the amount of water inside, then the condition inside will be above triple point after the steam being injected. Let's consider that the temperature of superheated steam is above 30C of its saturation level and the amount is far less than the amount of water inside.

Are you planning to model this step as a constant-pressure process? Adding mass will have to increase pressure as well for a fixed-volume container.

No. The pressure inside can't be the same after the superheated steam being injected. Both temperature and pressure inside will rise until an equilibrium point will come when all the steam inside will become saturated.

What you're describing sounds a little like a flash evaporation process in a vapor-liquid separator.

Again No. This is just opposite to flash evaporation. In flash evaporation process, pressure above water/liquid is suddenly reduced and a part of the liquid/water will become steam. In this process, the pressure will rise and part of steam will be added to the water inside so that its temperature can rise.

 

[Chestermiller]

No problem!

I guess when you said "No problem," what you really meant was "Yes problem." All I asked from you in post #10 was 6 numbers. After all the thinking you have done on this problem, it is hard to imagine that you would have trouble coming up with these 6 numbers.

 

[pranj5]

Initial Conditions in Tank
1. Volume of tank
2. Temperature in tank
3. Mass fraction of liquid water in tank

1. 1 cubic meter
2. 20C
3. 1 kg

Conditions of Superheated Injectate

1. Temperature
2. Pressure
3. Mass of superheated steam injected into tank

1. 60C
2. 7.37 kPa (saturation pressure of water at 40C)
3. 10 gm

 

[Chestermiller]

1. 1 cubic meter
2. 20C
3. 1 kg

1. 60C
2. 7.37 kPa (saturation pressure of water at 40C)
3. 10 gm

Good. Now, from the steam tables, the specific volume of saturated liquid water at 20 C is 0.001002 m^3/kg and the specific volume of saturated water vapor at 20 C is 57.79 m^3/kg. So, can you please calculate what part of the 1 cubic meter tank volume is occupied by the liquid water (in cubic meters)? What part of the 1 cubic meter tank volume is occupied by water vapor (in cubic meters)? What is the mass of the water vapor in the tank under these initial conditions? What is the total mass of water (both liquid water and water vapor) initially in the tank?

 

[pranj5]

Well, if the amount is 1 kg, then certainly the volume occupied by water is 0.001002 cubic meter and rest 0.998998 cubic meter has been occupied by saturated steam at 20C (Assuming the pot is very thin and its volume is negligible). As per steam table, the specific density of steam at 20C is 0.01728 kg/m^3 and therefore the mass of steam occupying 0.998998 cubic meter is 0.01726268544 kg. Therefore, the total mass of water and steam inside the tank initially was 1.01726268544 kg.

 

[Chestermiller]

Well, if the amount is 1 kg, then certainly the volume occupied by water is 0.001002 cubic meter and rest 0.998998 cubic meter has been occupied by saturated steam at 20C (Assuming the pot is very thin and its volume is negligible). As per steam table, the specific density of steam at 20C is 0.01728 kg/m^3 and therefore the mass of steam occupying 0.998998 cubic meter is 0.01726268544 kg. Therefore, the total mass of water and steam inside the tank initially was 1.01726268544 kg.

Excellent. So the mass of saturated liquid water in the tank is 1.000 kg and the mass of saturated water vapor in the tank is 0.0173 kg. Here is an excerpt from a more complete steam table than the one is Wiki for saturated steam:

It shows internal energy, enthalpy, and entropy of saturated water (for both liquid and water vapor) relative to liquid water at 0 C and 1 atm pressure. From the table you can see that the internal energy of saturated liquid water at 20 C is $u_L=$83.94 kJ/kg, and the internal energy of saturated water vapor at 20 C is $u_V=$2403 kJ/kg. Now, given the amounts of liquid water and water vapor in our tank initially, I would like you to calculate the total internal energy of the water in the tank initially.

Also, for later reference, these same more complete steam tables give a value of $h_v=2610$ kJ/kg for the enthalpy of suprerheated water vapor at 60 C and 7.37 kPa.

Now for the most important part. Are you familiar with the "open system" control volume version of the first law of thermodynamics (given in every serious thermodynamics text)? This version of the first law of thermodynamics applies to systems like ours in which mass is entering the system between the initial and final states of the system. If you are familiar with the open system version of the 1st law, please write the equation down. Thanks.

Chet

 

[pranj5]

The version just says that the mass at the beginning and at the end and also the gross amount of energy at the beginning and at the end will be same. I can't write the formula here as I don't know how to insert symbols in the thread. Whatsoever, you can continue.

 

[Chestermiller]

I guess you didn't want to calculate the total internal energy of the water and steam in the tank prior to the superheated steam injection (even though I requested that you do this). If you are uncomfortable with doing this calculation, I can do it for you. We will need the initial internal energy later in our calculations. Do you prefer that I evaluate the initial internal energy myself?

The version just says that the mass at the beginning and at the end and also the gross amount of energy at the beginning and at the end will be same.

This is not precise enough for what we are doing. We need to identify the precise quantities in the various terms of the equation.

Here is the open system control volume version of the First Law of Thermodynamics that is presented in Fundamentals of Engineering Thermodynamics by Moran, shapiro, Boettner, and Bailey (note that this was derived directly from the usual closed system version of the First Law):

$$\frac{dE_{cv}}{dt}=\dot{Q}_{cv}-\dot{W}_{cv}+\sum_i{\dot{m}_i(h_i+\frac{V_i^2}{2}+gz_i})-\sum_e{\dot{m}_e(h_e+\frac{V_e^2}{2}+gz_e})$$where the total energy in the control volume $E_{cv}$ is given by:$$E_{cv}=\int_V{\rho(u+\frac{V^2}{2}+gz})dV$$
In these equations, the subscript i refers to inlet streams to the control volume and e refers to the exit streams from the control volume. In addition h represents enthalpy per unit mass, u represents internal energy per unit mass, V represents velocity, g is the acceleration of gravity, z represents elevation above a specified datum, $\dot{m}$ represents the mass flow rate of an inlet stream or exit stream from the control volume, $Q_{cv}$ represents rate of heat addition to the control volume, and $\dot{W_{cv}}$ represents the rate of doing work on the boundary of the control volume, excluding the work to push material into and out of the control volume (the latter is included in the summation terms for the inlet and exit streams).

I think I'll stop here and give you a chance to digest these (standard) relationships before proceeding to show how they are to be applied to our specific problem.

 

[pranj5]

As per the list you have given, heat content in 1 kg of water at 20C is 83.94 kJ and that of 0.01726268544 kg of saturated steam occupying the volume is 41.48 kJ. In total, the gross heat content inside is 125.42 kJ.

 

[Chestermiller]

As per the list you have given, heat content in 1 kg of water at 20C is 83.94 kJ and that of 0.01726268544 kg of saturated steam occupying the volume is 41.48 kJ. In total, the gross heat content inside is 125.42 kJ.

Yes. You can call it the "gross heat content" if you want, but, in thermodynamics terminology, it is internal energy of the combined liquid water and steam in the tank initially $U_{initial}$, relative to the same mass of liquid water at 0 C and 1 atm.

 

[pranj5]

OK. What's the next step.

 

[Chestermiller]

OK. What's the next step.

The next step is to apply the open system version of the First Law of Thermodynamics to our particular system. Smith and van Ness, in Introduction to Chemical Engineering Thermodynamics, solve a very similar problem to ours in Chapter 7. Here's part of their analysis:
"If the tank is chosen as the control volume, there is but one opening into the tank and it serves as an entrance, because gas (in our case vapor) flows into the tank...In the absence of any specific information, we assume that kinetic- and potential-energy changes are negligible."

So, with these considerations, our open system First Law of Thermodynamics equation reduces to:
$$\frac{dU}{dt}=\dot{Q}_{cv}-\dot{W}_{cv}+\dot{m}_{in}h_{in}$$where U is the internal energy of the tank contents. Since the tank is rigid, $\dot{W}$ is equal to zero, and, since the tank is insulated, $\dot{Q}_{cv}$ is also equal to zero. So our equation reduces to:
$$\frac{dU}{dt}=\dot{m}_{in}h_{in}$$
If we integrate this equation with respect to time t from the initial thermodynamic equilibrium state to the final thermodynamic equilibrium state of our system, we obtain:
$$U_{final}=U_{initial}+m_{in}h_{in}$$where $U_{initial}=125.42$ kJ, $m_{in}=0.01$ kg, and $h_{in}=2610$ kJ/kg. So, the final internal energy of our system is:$$U_{final}=151.52kJ$$And the final mass of water in our system is $$m=1.0173+0.01=1.0273kg$$

Now, even though we know the final internal energy and the final mass of our system, we still don't know the final temperature and we don't know the final split between liquid water and steam. Determining these will be our next task.

I think I'll stop here for now.

 

[pranj5]

Now, even though we know the final internal energy and the final mass of our system, we still don't know the final temperature and we don't know the final split between liquid water and steam. Determining these will be our next task.

Now, point is whether the amount of water will increase or decrease. As per common sense, the amount of water should increase and I want to see whether that can be supported by calculation or not.

 

[Chestermiller]

At least, it's clear that a part of steam has been added to the water and that's why it's mass has been increased along with its temperature.

This seems correct. Now let's see by how much.

We basically have 3 equations that we need to solve simultaneously for the final temperature T, the final mass of liquid water $m_L$, and the final mass of water vapor $m_V$:$$m_L+m_V=m=1.0273 kg\tag{1}$$
$$m_Lv_L(T)+m_Vv_V(T)=V=1m^3\tag{2}$$
$$m_Lu_L(T)+m_Vu_V(T)=U_{final}=151.52kJ\tag{3}$$where $v_L(T)$, $v_V(T)$, $u_L(T)$, and $u_V(T)$ are the values of the saturated liquid specific volume, the saturated vapor specific volume, the saturated liquid specific internal energy, and the saturated vapor specific internal energy, respectively, at the final temperature T. The values of these parameters are to be evaluated from the steam tables at temperature T.

Eqn. 1 just represents a mass balance on our system. Eqn. 2 is a constraint that requires that the total final volume of liquid and vapor in the tank is equal to the tank volume. Eqn. 3 is the result of our open system First Law of Themodynamics energy balance.

To solve Eqns. 1-3, we define x as the fraction of the final tank contents that is vapor. Then the fraction of the final tank contents that is liquid is (1-x). So, the final mass of liquid in the tank is $$m_L=m(1-x)\tag{4}$$ and the final mass of vapor in the tank is $$m_V=mx\tag{5}$$. Eqns. 4 and 5 satisfy our overall mass balance equation (Eqn. 1) identically.

If we substitute Eqns. 4 and 5 into Eqns. 2 and 3, we obtain:
$$v_L(T)+(v_V(T)-v_L(T))x=\frac{V}{m}=\bar{v}\tag{6}$$
$$u_L(T)+(u_V(T)-u_L(T))x=\frac{U_{final}}{m}=\bar{u}\tag{7}$$
where $\bar{v}=0.9734\ m^3/kg$ and $\bar{u}=147.49\ kJ/kg$. We can now eliminate the mass fraction vapor x from Eqns. 6 and 7 to obtain a relationship solely in terms of the final temperature T:
$$\frac{[0.9734-v_L(T)]}{[v_V(T)-v_L(T)]}=\frac{[147.49-u_L(T)]}{[u_V(T)-u_L(T)]}\tag{8}$$
At the final temperature, the right hand side of this equation must equal the left hand side. Thus, if we write
$$f(T)=\frac{[0.9734-v_L(T)]}{[v_V(T)-v_L(T)]}-\frac{[147.49-u_L(T)]}{[u_V(T)-u_L(T)]}$$
we must solve for the value of T for which f(T) is equal to zero.

Pranj5: I am requesting that you determine the values of the function f(T) at 20 C and 30 C using the section of the steam tables that I presented in my earlier post. I think that the final temperature will lie in this range.

 

[pranj5]

At 20C,

f(T)

=[{0.9734-1}/{0.999-1}] - [{147.49-83.94}/{2403-83.94}]=-1.0266-83.91355= -84.94.
Will give information about 30C later.

 

[Chestermiller]

At 20C,
=[{0.9734-1}/{0.999-1}] - [{147.49-83.94}/{2403-83.94}]=-1.0266-83.91355= -84.94.
Will give information about 30C later.

The numbers listed in the table for specific volumes of the liquid are the actual values multiplied by 1000. Also, the specific volume of the vapor at 20 C is not 0.999. It is 57.79 m^3/kg. The numbers you show for the 2nd fraction (involving internal energies) are correct, but the calculated value of the fraction should only come out to be 0.0274 (not 83.91).

 

[Chestermiller]

I just noticed that the steam table excerpt I gave only goes up to 28 C, so please evaluate f(T) at that temperature rather than 30 C. Sorry.

 

[pranj5]

The numbers listed in the table for specific volumes of the liquid are the actual values multiplied by 1000. Also, the specific volume of the vapor at 20 C is not 0.999. It is 57.79 m^3/kg. The numbers you show for the 2nd fraction (involving internal energies) are correct, but the calculated value of the fraction should only come out to be 0.0274 (not 83.91).

Sorry, my mistake! I have thought that the actual volume of steam in this case should be the number instead of specific volume.

 

[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?