Question about the latent heat of vaporization of water

[adlep]

The latent heat of vaporization of water at 1atmospere is calculated to be 2260.44J/g
So it takes 540J of energy to break the strong bonds of hydrogen atoms and convert the water into a steam. This is at the regular 100 degrees of C.

Now lets assume that we are in the Himalayas. The atm pressure at the altitude of 8000M is only about .3 of the atm pressure at the sea level.
The boiling point of water at 8000m moves down from 100 degrees C to 75.5C

Question - will the lower pressure also affect the latent heat of vaporization constant?

I sound to me that it should be lower as the pressure decreases. Because of the lower pressure, less force is required to break the compressed bonds of the hydrogen atoms.

Please elaborate.
?


Edit: I also have additional question.
How much energy is required to distill the 1 liter of salt water into fresh water at 8000m?

Assuming the boiling point of water is 75.5C
Also,
I would assume that at least some fraction of the energy that was used to boil the water, comes back into the system as the water vapor cools down and releases that extra energy.

 

[Paperdoc]

I can't answer your first question for sure, but I would EXPECT that the Heat of Vaporization is relatively independent of the surrounding pressure, especially over the range from 1.0 to 0.3 atm.

On the additional question, capturing the heat released is a practical problem. You add enough heat in the boiler chamber to a liquid at 75.5C to convert it to a gas. The gas travels from boiling chamber to condenser and there must release the heat of vaporization at 75.5C to some heat absorber. The absorber must be cooler than 75.5C for this to happen. So now you have heat captured in a medium (air or water) used for cooling the condenser, at a temperature of less than 75.5C. How do you get that energy back into the boiling chamber at 75.5C? I can speculate that, in a larger efficient system, you might design with a heat pump system to cool the condenser with a refrigerant, then compress it and pump it through the boiling chamber at higher temperature to release the heat there. However, I do not know whether the energy efficiency of such a design would warrant the captital and operating costs of the heat pump system, versus the simple system of burning fuel under the boiler (or using electrical heating) and air or cold water for condenser cooling.

One very common way to use SOME of the heat at the condenser, though, is cheap and simple. You use the raw cool feedstock to the system (in this case, salt water) as the cooling medium feed to the condenser, and control its flow to the minimum necessary to ensure stable operation and condensing of all vapor. The heated cooling water then is sent to the boiler, but only partly. The flow into the boiler will be less than the flow through the condenser, but at least it is pre-heated and you have captured some of the vapor's heat for re-use in the system. What heat is lost is that contained in the portion of the condenser cooling water exit that cannot be used in the boiler. The major trick to recognize here, though, is that using salt water as the condenser cooling medium imposes constraints on that condenser design. Now it MUST be constructed of a material that will withstand constant exposure to moving salt water at high temperature without corroding away rapidly. Those same constraints are part of the boiler system design, but are not necessary at the condenser if you choose a different cooling supply. The additional capital cost of the condenser, plus its anticipated maintenace / replacement costs over the system lifetime, must be balanced against the operating cost savings available through reduced fuel consumption at the boiler.

 

[CycloWizard]

It's been a while since I took thermo, but I think I can still field this one. The latent heat does depend on pressure because the equilibrium pressure and temperature are related. Looking back into my thermo book, I see that the latent heat is equal to T*delta(V)*dP_sat/dT for a pure species (this is known as the Clapeyron equation, which is derived in many chemical engineering thermo books at least).

Looking at some diagrams in the appendix, it appears that the latent heat generally decreases as pressure increases. My guess is that this is because of the changing nature of molecular interactions as the "fluid" nears its triple point, above which solids can actually evaporate. Essentially, since equilibrium occurs at lower temperatures at elevated pressures, different bonding processes dominate than do at higher temperatures.

I can also say with certainty that the rate of evaporation depends critically on the pressure. While the vapor pressure depends on temperature only, the driving force for evaporation is a gradient in partial pressure. Partial pressure is the mole fraction of the evaporating species divided by the total pressure, and will therefore obviously be smaller far from the "puddle" at higher pressures, creating a larger gradient.

As for the second question, it's a bit complicated because the concentration of salt in the water is a key variable. I believe that normal seawater is just over 3 wt% salt. I'd have to do some digging for a T(x,y) diagram at the right pressure to get a ballpark answer or some data on saltwater to calculate it, but I don't really have time right now. It can certainly be done though.

As for the very last part, energy from the steam will only return to the boiling pot if the steam condenses. That energy now "belongs" to the steam and is necessary to keep it in the vapor phase. This is why liquids actually cool down as they evaporate. For example, you sweat because the evaporating sweat cools you down. Boiling water keeps boiling because you keep adding heat to it. As soon as you remove it from the heat source, it stops boiling because there is no source of energy, but energy is being released as the water vaporizes.

 

[Veramocor]

Question 2: Key words: Multi-effect Evaporation, this is one method by which desalination plants can work,


The amount of energy to distill the water really depends on how much capital you want to spend. If you only had 1 heat exchanger (one effect), you would need about 1 pound of steam to evaporate 1 pound of water. If you had 2 effects you'd evaporate 1.6 lbs of water. Three effects, you evaporate 2.6 lbs of water. A good rule of thumb is # effects * 0.8 = the steam economy or lbs of water evaporated per lb of steam used.

How does it work?

Three effect evaporator:

Steam will flow from 1st effect ->2nd -->3rd --> a condenser
The salt water will flow from 3rd-->2nd-->1st

Lets say you have a boiler that produces 65 lb steam. You send the 65 lb steam to the 1st effect, the steam condenses and gives its heat of condesation to stalt water flowing to it. This causes the salt water to partially evaporate. The steam from the evaporated salt water is then sent to the second effect, which in turns evaporates the part of the salt water going to it. The generated steam then goes to the third effect. Finally the steam from the third effect goes to the condenser. The salt water gets more and more concentrated as it goes from the third to the 1st effect till you have a concentrated brine. The condensates you collect are the distilled water.

The delta T between the steam and the condenser cooling water helps determine how big each effect has to be. A final interesting point is as the salt solution becomes more saturated you have to deal with boiling point rise. Which means that at a given salt concentration you need a higher temperature to evaporate the solution then you would need to evaporate pure water.

 

[ShawnD1]

Originally posted by: CycloWizard
It's been a while since I took thermo, but I think I can still field this one. The latent heat does depend on pressure because the equilibrium pressure and temperature are related. Looking back into my thermo book, I see that the latent heat is equal to T*delta(V)*dP_sat/dT for a pure species (this is known as the Clapeyron equation, which is derived in many chemical engineering thermo books at least).

I'm very certain this is wrong. The Clausius-Clapeyron equation is supposed to show a relation between temperature and vapor pressure; latent heat is only in the equation because it's a constant that completes the equation. If you're an engineer, the term "fudge factor" accurately describes what the latent heat value is.

Latent heat is just another form of enthalpy, and enthalpy is temperature-independent.


You might be thinking of entropy and how it relates to Gibbs free energy. Gibbs free energy is a way of calculating whether or not a change will happen to a system, and it's calculated by subtracted the entropy change from the enthalpy change. Entropy is where most of your attention should go when looking at free energy since entropy is heavily influenced by both temperature and pressure.

Quote from Wiki:

Ice melting example

Main article: disgregation

The illustration for this article is a classic example in which entropy increases in a small 'universe', a thermodynamic system consisting of the 'surroundings' (the warm room) and 'system' (glass, ice, cold water). In this universe, some heat energy dQ from the warmer room surroundings (at 298 K or 25 C) will spread out to the cooler system of ice and water at its constant temperature T of 273 K (0 C), the melting temperature of ice. The entropy of the system will change by the amount dS = dQ/T (this means the enthalpy of the reaction steals energy from the room in the form of entropy), in this example dQ/273 K. (The heat dQ for this process is the energy required to change water from the solid state to the liquid state, and is called the enthalpy of fusion, i.e. the ?H for ice fusion.) The entropy of the surroundings will change by an amount dS = -dQ/298 K. So in this example, the entropy of the system increases, whereas the entropy of the surroundings decreases.

Basically what it's saying is that melting ice (or boiling water) is a change where enthalpy is trying to stop the change from happening. The change happens anyway because the amount of entropy present is greater than the enthalpy of that particular change. In very simple terms, an environment with a higher temperature has more entropy, so an ice cube will melt faster in a hotter room. When talking about gases, entropy increases as pressure decreases. If more of the entropy can be taken from having a lower air pressure, the amount of entropy taken from temperature can be less. Although water can boil at 20C instead of 100C when the pressure is dropped, the overall change of entropy is the same; what changes is where that entropy came from (pressure rather than temperature).

 

[CycloWizard]

Originally posted by: ShawnD1
I'm very certain this is wrong. The Clausius-Clapeyron equation is supposed to show a relation between temperature and vapor pressure; latent heat is only in the equation because it's a constant that completes the equation. If you're an engineer, the term "fudge factor" accurately describes what the latent heat value is.

No. The Clausieus-Clapeyron equation is different than the Clapeyron equation. The former is an approximation of the latter that adds two limiting assumptions:
1. The vapor phase behaves as an ideal gas, and
2. the molar volume of hte liquid is negligible compared with the molar volume of the vapor.
The CC equation then has the form latent heat=-R*d(ln(P_sat))/d(1/T), implying that latent heat is almost independent of temperature, which is often a poor approximation except at very low pressures (i.e. where ideal gas assumptions hold). The Clapeyron equation is actually exact and can be derived directly from the fact that the Gibbs free energy of the liquid and vapor phases are equal at equilibrium. Finally, latent heat is NOT a "fudge factor" - it is a measurable quantity that is also very predictable given knowledge of the composition and conditions that you're interested in. I'm not quoting from Wiki, I'm paraphrasing from Smith, Van Ness, and Abbott (2001): Introduction to Chemical Engineering Thermodynamics, 6th edition, pp. 217-218 (which is the Bible of thermodynamics, at least for chemical engineers). Stand back - I've been training for this my whole life!

 

[ShawnD1]

My mistake. Most people don't cite the Clapeyron equation for everyday thermodynamics since it adds a lot of needless complexity in atmospheric conditions. I incorrectly assumed that you were trying to recite something from a class you took recently, such as last year.

I also do this for a living, but I'm a research chemist rather than an engineer. From a research standpoint, most of this stuff is done in a very half-ass fashion and everything is put in terms of ratios and fudge factors.If I oversimplified thermodynamics, or put down the complexity of your work, I apologize.

 

[CycloWizard]

Originally posted by: ShawnD1
My mistake. Most people don't cite the Clapeyron equation for everyday thermodynamics since it adds a lot of needless complexity in atmospheric conditions. I incorrectly assumed that you were trying to recite something from a class you took recently, such as last year.

I also do this for a living, but I'm a research chemist rather than an engineer. From a research standpoint, most of this stuff is done in a very half-ass fashion and everything is put in terms of ratios and fudge factors.If I oversimplified thermodynamics, or put down the complexity of your work, I apologize.

It has been a while since I actually took a thermo class, but I've more recently taught a class in thermo with this guy, who is one of the godfathers of thermodynamics. I don't do thermo research, nor am I offended by what you said, but what you said certainly doesn't apply to much (if any) of the chemical engineering community, particularly the research community.

 

[adlep]

Originally posted by: Veramocor
Question 2: Key words: Multi-effect Evaporation, this is one method by which desalination plants can work,


The amount of energy to distill the water really depends on how much capital you want to spend. If you only had 1 heat exchanger (one effect), you would need about 1 pound of steam to evaporate 1 pound of water. If you had 2 effects you'd evaporate 1.6 lbs of water. Three effects, you evaporate 2.6 lbs of water. A good rule of thumb is # effects * 0.8 = the steam economy or lbs of water evaporated per lb of steam used.

How does it work?

Three effect evaporator:

Steam will flow from 1st effect ->2nd -->3rd --> a condenser
The salt water will flow from 3rd-->2nd-->1st

Lets say you have a boiler that produces 65 lb steam. You send the 65 lb steam to the 1st effect, the steam condenses and gives its heat of condesation to stalt water flowing to it. This causes the salt water to partially evaporate. The steam from the evaporated salt water is then sent to the second effect, which in turns evaporates the part of the salt water going to it. The generated steam then goes to the third effect. Finally the steam from the third effect goes to the condenser. The salt water gets more and more concentrated as it goes from the third to the 1st effect till you have a concentrated brine. The condensates you collect are the distilled water.

The delta T between the steam and the condenser cooling water helps determine how big each effect has to be. A final interesting point is as the salt solution becomes more saturated you have to deal with boiling point rise. Which means that at a given salt concentration you need a higher temperature to evaporate the solution then you would need to evaporate pure water.

Folks I really appreciate your feedback.
Also,
Again, is there any chance for a specific value of the latent heat of vaporization at 8000m?

Again, it is calculated at 2260.44J per each gram of water at the sea level.

 

[Veramocor]

http://www.efunda.com/Material...ter/steamtable_sat.cfm, No need to mess with equations the work has been done already. Type in pressure or temp of vaporization.

 

[CycloWizard]

Originally posted by: adlep
Folks I really appreciate your feedback.
Also,
Again, is there any chance for a specific value of the latent heat of vaporization at 8000m?

Again, it is calculated at 2260.44J per each gram of water at the sea level.

For salt water or pure water? If salt water, what salt concentration?

 

[miniMUNCH]

Enthalpy is state function meaning that it doesn't matter how you get to a state.

0.3 atm water ---compress-> 1 atm - water ------Vaporize------> 1atm - steam ---decompress---> 0.3 atm- steam

So the difference in the enthalpy of vaporization of water at 0.3 and 1.0 atm is equal to:

[dH/dP]water - [dH/dP]steam (over the pressure range 0.3 to 1.0)

since the change in enthalpy due to decompression of a gas in larger, in general, than that of the liquid the heat of vaporization will be less at 0.3 atm than at 1.0 atm.

In terms of macro thermo, the above explanation is 'why'


cheers

 

[jagec]

Originally posted by: Veramocor
Question 2: Key words: Multi-effect Evaporation, this is one method by which desalination plants can work,


The amount of energy to distill the water really depends on how much capital you want to spend. If you only had 1 heat exchanger (one effect), you would need about 1 pound of steam to evaporate 1 pound of water. If you had 2 effects you'd evaporate 1.6 lbs of water. Three effects, you evaporate 2.6 lbs of water. A good rule of thumb is # effects * 0.8 = the steam economy or lbs of water evaporated per lb of steam used.

How does it work?

Three effect evaporator:

Steam will flow from 1st effect ->2nd -->3rd --> a condenser
The salt water will flow from 3rd-->2nd-->1st

Lets say you have a boiler that produces 65 lb steam. You send the 65 lb steam to the 1st effect, the steam condenses and gives its heat of condesation to stalt water flowing to it. This causes the salt water to partially evaporate. The steam from the evaporated salt water is then sent to the second effect, which in turns evaporates the part of the salt water going to it. The generated steam then goes to the third effect. Finally the steam from the third effect goes to the condenser. The salt water gets more and more concentrated as it goes from the third to the 1st effect till you have a concentrated brine. The condensates you collect are the distilled water.

The delta T between the steam and the condenser cooling water helps determine how big each effect has to be. A final interesting point is as the salt solution becomes more saturated you have to deal with boiling point rise. Which means that at a given salt concentration you need a higher temperature to evaporate the solution then you would need to evaporate pure water.

And let me tell you, multi-effect evaporators are a bear to get into steady state. Startup procedures are touchy and it's easy to flood a stage, or boil it dry, without very careful control on a bunch of linked variables. Apparently in paper mills, the startup procedure can take a week before the thing's running at steady state...and the engineers who run them would sell their firstborn to avoid having to shut them down for any reason once theyr'e up.