The three vessels will all be at the same temperature as the surroundings. If the solution in the vessel is a pure component it is likely the material will be 100% liquid or 100% vapor depending on whether the surrounding temperature is above or below the boiling point of the material in the vessel. If it is a mixture with some components boiling bellow the ambient temperature and some above it will be mixed phase at the vapor composition. If the ambient is cold enough it ill be 100% liquid or even solid if it is hot enough it could be 100% vapor in between the contents will be vapor/liquid and/or solid liquid equilibrium. If it is a closed vessel pressure will increases as the contents vaporizes or decrease as it condenses and a new VLE will be achieved at the higher or lower pressure.

Saturation temperature will depend on the composition of the material in the vessel and the steady state pressure achieved as that material vaporizes or condenses to the steady state phase equilibrium.

Dear Dr. Aouf, As Mr. Rick Manner described above, for single component in mixture I mean if there are no impurities in mixture and any other gases with vapor pressure, the saturation temperature for three vessel even if with different amount of vapor-liquid mixture will be same saturation temp. you can look at Figure 8.4 in this link to see more explanation under temp Vs volume for mixture state. I think pressure system and mixture components more influence parameters on saturation temperature of mixture in thermal equilibrium system. Regards.

http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node61.html

Dear Dr. Momammed saad Kamel

Thank you for your response in answering my question, however, I do think the saturation temperature will be different if different amounts of fluid are in each vessel. I think the saturation temperature is a function of the amount of fluid since increasing the amount of fluid will increase the pressure which in turn increases the saturation temperature.

Aouf,

Yes for each component Ki = Yi/Xi = VPi/P so as pressure changes % vaporized will change if the mixture is in a vapor-liquid two phase region and yes there may be small differences between VP and fugacity and non unity activity factors for the liquid so Ki may not be exactly VPi/P but it will probably be pretty close.

Aouf Al-Tabbakh,

If you have a pure component, it is possible to have three containers one with 10% vapor, one with 50% vapor and one with 90% vapor. If this is, say water and the pressure is atmoshperic then the temperature in all containers will be 212°F. That's the saturation temperature. If the environment temperature is 212, then the three systems will sit there unaffected. If it was n-pentane, then the saturation temperature for all conatiners will be 97°F. If the environment temperature is 97, then the three systems will sit there unaffected. Since the containers have the same volume, then the total mass in each container will be different if all containers are to have the same pressure.

Take a look at the phase rule see if it can help you out.

It's enough to know what saturated vapour is. Regards,

Thank you for all the respected participants in answering my question. However, I just want to focus on the fact that:- the temperature in the three containers will be the same and equals the surrounding temperature, but the saturation pressure would not necessarily be the same. Each vessel would require different amount of heat to reach its own saturation pressure and starts the boiling. I do not think that the difference in mass will keep the same saturation pressure. I still need more precise answer, please.

Dear Dr. Aouf Al-Tabbakh

1- What is the saturation temperature of a liquid-vapor mixture in a sealed vessel in thermal equilibrium with surroundings? Is it the surrounding temp.?

[Ans] The saturation temperature of a liquid-vapor mixture that is in thermal equilibrium with the surrounding must be equal to the surrounding temperature because thermal equilibrium means that the temperatures are equal and the zero law of thermodynamics states that "If two body temperatures are equal to the temperature of a third body then the temperature of these two bodies are equal".

2- Suppose you have 3 vessels of equal volume and equal temperature but containing different amounts of liquid-vapor mixtures. If the 3 vessels are in thermal equilibrium with environment, what is the saturation temperature of each vessel?

[Ans] The saturation temperature of each mixture in any of the 3 vessels (containing different amounts of liquid-vapor mixtures) that is in thermal equilibrium with the surrounding must be equal to the surrounding temperature as proved in part 1 above leading to the same value of saturation pressure in any of these vessels. But each vessel will have different value of dryness fraction (x). Example: (vessel 1 = 10% liquid , vessel 2 = 50% liquid and vessel 3 = 90% liquid) then (Dryness fraction "x" of vessel 1 = x1= 90%, x2 = 50% and x3 = 10%).

Aouf Al-Tabbakh - please clarify you question speifically:

* is it a single component or multicomonent system.

* What is the starting condition

- do you fill and partially drain or is there air/nitrogen in the vessel

- Are the starting condition such that it is 100% vapor/liquid or solid or mixed phase

Dear Dr. Rick Manner..

Thank you for your interest in answering my question..

1- The three vessels contain single component.

2- The masses in the vessels are different (say; 1 kg, 2 kg, 3kg).

3- The volume is the same for the three vessels.

4- The three vessels are initially evacuated from air then charged with different amounts of a certain refrigerant.

5- The charging is performed at the same temperatures for the three vessels. The vessels are then sealed and left to reach thermal equilibrium with environment.

(( I do think that if we measure the pressure (not saturation pressure) in the three vessels it would be different in each vessel. As a result, the saturation pressure will also be different in each vessel, meaning that each vessel requires different amount of heat to commence boiling and change the vapor-liquid ratio))

Am I correct in what I postulate?

Thanks and best regards.

Dear Dr. Mishaal A. Abdulkareem..

Thank you for your interest in answering my question.

If what you mentioned is true, what will happen if we supply each vessel with equal amount of heat and perfectly insulate them. The temperature in each vessel will increase at different level due to the difference in vapor-liquid content in each vessel. So, different amount of liquid in each vessel will convert to vapor causing different pressure rise in each vessel. Eventually the saturation temperature will be different in each vessel. So the difference in the amount of fluid causes a difference in saturation temperature when equal amounts of heat are supplied. The case of equal saturation temperatures in the three vessels occurs only when thermal equilibrium with environment is established. Otherwise the saturation temperature will vary in different manner for each vessel even if same boundary conditions are varied for the three vessels.

@ Aouf Al-Tabbakh

What will happen if we supply each vessel with equal amount of heat then perfectly insulate them ?

[Ans] Since all these vessels were initially (state 1) in thermal equilibrium with the environment and they were at the same value of saturation pressure and saturation temperature but with different values of dryness fraction, then the same value of heat is added to each vessel, then all vessels were perfectly insulated. This is called "constant volume heat addition process" and all vessels will receive the same value of latent heat of evaporation. Evaporation will starts and continue (During heating process) in all vessels and the value of dryness fraction in each vessel will increase in the same speed (at this point all vessel are still at the same initial value of sat. press. and sat. temp.) until the vessel that was initially with the higher value of dryness fraction (vessel 1) will be empty of its liquid (will be fully dry >> x=100%) then its dry vapor will start to superheat and the superheated vapor will start to behave as gas and it will follow the perfect gas law (PV=mRT). Since it is a constant volume heat addition process, then the temperature will increase [Q=mCv(T2-Tsat.), T2>Tsat.] and as a result the pressure will increase [P2/T2=Psat./Tsat.=mR/V=constant >> P2=Psat.(T2/Tsat)].

The temperature of vessel 1 will continue to increase above saturation temperature as long as heat is added and this will cause an increase in pressure above its saturation value at the same time other vessels will stay at its initial values of sat. temperature and pressure until they will be fully dry (x2 or x3 = 100%) then an increase of temperature will start with the vessel that was dried first.

In general, the increase in temp. and press. will start in any vessel immediately as soon as fully dryness is reached first.

Dear Dr. Mishaal

The explanation you added is correct from thermodynamics point of view. However, the difference in mass in each vessel means different volume occupied by vapor in each vessel, hence, the generated vapor quantity, even if it is equal for the three vessels would not found equal volumes to occupy. This will lead to a difference in saturation pressure in each vessel. The vessel with least dryness fraction (highest liquid quantity) will have the least vapor volume and therefore will have the highest increase in pressure among the three vessels for the same amount of heat added to each vessel.

@ Aouf Al-Tabbakh

The generated vapor quantity in each vessel, will occupy a volume that was occupied when it was liquid before evaporation. The decrease in liquid volume in each vessel due to its evaporation must equal the increase in vapor volume after evaporation.

@ Mishaal A Abdulkareem

The generated vapor can not occupy the same volume it occupied when it was liquid. Its volume increases due to inherent nature of vaporization. The limited volume available for expansion is then causes an increase in pressure. This phenomenon can be tested by pressurized cooking pots. When the pressure inside the pot increases due to the extra amount of generated vapor the relief valve opens and discharge some vapor to stabilize the pressure inside the pot.

Aouf,

If it is a single component there are several possible cases including:

100% Vapor Phase

Mised phase vapor and liquid

100% liquid phase

mixe solid and liquid

100& Solid phase

I think you intend for the conditions to be such that it is mixed phase vapor and liquid. This particular system is defined by a few fundamental equations:

total wessel volume = volume of liquid + volume of vapor

K(t) = y/x = vp(t)/P (three might my a small correcton with an alpha)

and

Volume of vessel= V liq + V vapor

Vliq = n liq *liq density (m3/mole)

V vap =

Dear Dr. Aouf Al-Tabbakh

The saturation Temp depends on the pressure in the vessel like the heat pipe deep vacuum lower saturation Temp. moreover the filling ratio of the working fluid.

Thanks and best regards.

Dear Dr. Adel A. Eidan

This is the point I wanted to reach:- Is there a relation between the initial charge inside a container (e.g. heat pipe) and the operating saturation temperature inside it?

@ Aouf Al-Tabbakh

Yes, my last answer needs some correction. and i know the correct answer and i will answer it after doing some shopping. thank you

@ Aouf Al-Tabbakh

When evaporation starts at constant volume, the generated quantity of dry vapor (vg) that is at an initial sat. temp. (Tsat.) and initial sat. press. (Psat.) will needs a volume that is much larger than its volume when it was sat. liquid (vf) ( i.e. vg >> vf ). That means its volume will be magnified (the magnification ratio = vg/vf). Since the available volume is vf, therefore the volume that is needed to occupy the rest of the generated dry vapor is not available = vg - vf. As a result, it will be compressed and a small increase in pressure will be observed that is the new sat. pressure which is higher than the initial value of sat. press. (and the new corresponding sat. temp. will be higher than the initial sat. temp.). this process will continue and the sat. press. will increase gradually and so do the sat. temp. and the more the sat. press. increase the lower is the magnification ratio (vg/vf) giving a smaller and smaller available volume to the new small amount of generated dry vapor volume. Therefore, the relation between Psat. and vg is not linear, instead its near power function.

This case is (lumped system) because it is assumed that all the domain have the same temperature and it is transient in nature and it needs a transient mathematical model that you are looking for.

Sketch a simple physical model then put all the variables that you need in your analysis then write the governing equation then write a code (subroutine) to predict all the thermodynamic properties of the liquid-vapor mixture of the fluid in question then link this subroutine to a main code (MATLAB or FORTRAN or any programing language that you prefer) that solves your problem.

Thank you Dr. Mishaal .. The conversation was really fruitful !

I will refer to the question once again.

Saturated steam, specifying - its pressure, over the pure component depends only on the temperature. This is the saturation temperature. The volume of liquid does not affect the value of the process parameters at equilibrium. Here, however, not only the volume of the liquid is different, but there is also a different volume of air over the pure component. Besides, the vessel is sealed. Will the same amount of liquid evaporate at the same temperature (for a sufficiently long time) ? The answer is no, because it would determine different partial pressures of the evaporated component. It is also worth remembering that the different amount of evaporated component will affect the relation of gas phase volume to liquid phase volume. Of course, the higher the temperature, the higher the impact.

In my opinion, in this situation, the concept of saturation temperature (as it is interpreted for the open system) loses meaning. Assuming no energy exchange resistance and perfect mixing in gas phase, the temperature in a sealed vessel might be considered as a so called saturation temperature. But what for. Regards,

Dear Dr. Miroslaw Grzesik

Thank you for response..

The sealed vessel in my case contains pure water without air (single component). Suppose we have three identical vessels containing different masses of pure water. If the three vessels are furnished with identical rates of heat, does the pure water in the three vessels boils at the same boiling (saturation temperature) knowing that the volume is the same for the three vessels?

In such a situation, after a sufficiently long time we will have to deal with the state of equilibrium: water - steam (liquid - vapour). At a constant outside temperature (constant heat rate), the temperature in each vessel will be the same equal to the saturation temperature and the pressure will be equal to the saturated vapor pressure. However, the time to achieve it will be different. Larger for a smaller volume of water in the vessel (constant volume of vessels). Of course, then there will be no difference between a closed and an open system. The boiling point should be the same. Regards,