Steam: getting past the old school thinking with patent 7588074

[1bobwhite]

Can anyone see the advantages of making use of the information contained in this patent. It seems industry has been old school thinking forever.

In my research for this patent, I made use of steam tables that were generated in the early years of the last century. The same tables are being cited in todays textbooks and writings. The problems started when examining the information derived from the tables.

 

Professors and others quoting them made assumptions that I believe were not correct, and then expounded on ways and methods, with elaborate diagrams and charts, to show how to interpret and correct the real world applications of this information.

 

For instance, when there is a problem of air in a steam system using a heat exchanger as the example, the assumptions were that the partial pressures of the air would reduce the efficiency of the heat exchanger because it lowered the steam temperature. They claimed that a 10% air volume would leave a 90% steam volume and with 100 psi pressure that would be steam at 90 psi.

Then they would cite from the steam tables, the saturated steam temperature for 90 psi steam as proof of the lower temperature, giving that as the reason for the lower temperature out of the exchanger.

 

First Google the term "partial pressures of steam" and read what comes up.

The first thing to come up for me is The Engineering toolbox with the "air and steam mixture" information. And they cite the Daltons Law of Partial Pressure which is; The total pressure exerted by a mixture of gases is the sum of the partial pressures of the individual gases!

There are several things wrong with this reasoning. First, the idea of partial pressures is for calculation purposes only and not for real world, because all the gasses within the vessel will be at the same pressure.

 

Then the partial or effective pressure of steam that they use is converted from percentage of volume into percentage of pressure. Here again the 90% steam by volume will not be at a different pressure within the vessel.

 

Then when they go to the saturated steam tables, they cite the 90 psi steam pressure to get the lower steam temperature. Magic. Voodoo actually, because the steam isn't saturated, its only 90%.

Therefore because of all of this misinformation, I decided to design a new approach to the concept of heat exchange. Hence the successful patent application and issue.

[insane_alien]

no, you're the one who has made some fundemental misunderstandings. you see, the calculations we do today, with the method you say is wrong, work. not only that, bthey work extremely well infact, i did a design calculation for a heat exchanger of a mixed air/steam saturated mixture and the temperature was correct to 3 decimal places.(the instrument i was using for temperature measurement did not measure any more accurately than that.

 

using your method, the calculation would be wrong by 2 orders of magnitude.

 

daltons law is fine. it is discussing the fraction of the pressure provided by one particular component of the gas. it works. we know it works. there is over a century of data proving it works.

 

you also mis understand the term saturated. saturated doesn't mean its 100% steam. it means that if you added any more steam and kept the temperature the same the steam would start to condense.

 

if you actually managed to get a patent on this thenthe patent office are even more retarded than i thought. especially as they shouldn't be patenting stuff like this. try submitting your idea to a journal. a chemical engineering journal or physical chemistry journal would be the most appropriate. i guarantee that it will not be published as it is bumf.

 

also, have you done any experiments to test this idea? do you have any data at all?

 

EDIT: i jsut looked up that patent http://www.freepatentsonline.com/7588074.html

 

it has bugger all to do with what you're talking about. its to do with pin finned heat sinks, nothing at all about an air-steam mixtures thermodynamic properties.

 

lies just dig the hole deeper. you're not going to manage to dig yourself out by going that way.

Edited May 9, 2010 by insane_alien

[1bobwhite]

I am really sorry that I touched a raw nerve. If you will please bear with me as I go through the explanations of my statements, it would really help.

 

First, I believe the numbers of the steam tables are good and accurate and are proven through all of the calculations.

 

On one issue; Dalton's Law, I wish for you to consider an illustration that I will present.

 

The image I will use is of a test arrangement that anyone may construct to demonstrate what I am about to describe.

 

It is a closed vessel or rigid envelope, inside of which are two chamber areas.

Chamber 1 area inside the vessel has attached to its incoming pipe, a loose collapsed bag that will contain gas #1, which will be carbon dioxide from your breath, the rest of the area is chamber 2 in which is gas #2 which is just plain air.

Pressure gauges are attached so that they will read each chambers pressure independently.

 

Now first, with both of the chamber areas closed off by the valves, and no other paths for the gasses to move, the pressure readings of both gauges will be the same. Chamber 1 volume will be a small percentage of the total, and chamber 2 volume will be the remaining.

 

Next, open the valve for chamber 1 and blow breath into its chamber. You will notice that the pressure of both gauges increase as you increase breath pressure. Chamber 1 volume has increased to a much higher percentage of the total chamber volume, with a corresponding decrease of chamber 2s volume, and both chambers gauges will read the same pressure.

 

Now, close off the valve so that the higher breath pressure remains in chamber 1.

 

Now just a few questions.

1. Is chamber 1 pressure the same as chamber 2 pressure?

2. Is chamber 1 volume a percentage of the total volume?

3. Is chamber 1 gas a different gas than chamber 2?

4 Does chamber 1 gas at its smaller percentage of volume have the same pressure as chamber 2 gas?

5. If the membrane of the bag was removed so that the gasses intermingled and uniformly blended, would there be any change in the overall contained pressure?

 

And one final question, since these two different gasses had the same pressures throughout the experiment, how can it be determined by experiment that they each have a different partial pressure?

 

Conclusion: Daltons Law cannot be demonstrated and verified by experiment.

 

However for calculation purposes, Daltons Law works very well for ideal gasses.

 

For steam though other factors are at work as I will go through in my next post.

[Mr Skeptic]

And how about that patent? We don't appreciate people making stuff up and pretending it is real. What does that patent have to do with this topic? Did you make it up, or did you get the wrong number?

[1bobwhite]

Mr Skeptic,

Yes it is my patent, and steam was my main concern for developing it.

Have you ever heard of "heat pipes". Do you know how they work?

 

The main premise for the patent was to take advantage of the fact that energies propagate back and forth through what is called the boundary layer at rates that have the inverse square rate of transfer. All energies of propagate in this manner.

 

My design for the exchanger is to make the fin radiating material also follow this inverse square rate as close as possible so that for instance the larger surface contact area of the fin would sink the heat the fastest because of its greater mass and the larger surface contact area and transfer that heat up into the fin at the fastest rate possible by conduction. As the heat is progressing up the fin it is also being cooled by the largest fin area at the base. The further up the fin, the lesser amount of heat has progressively less and less surface area to ideally match the amount of heat left to transfer, until approaching ambient, when it becomes uneconomical to have any more pin material. This entire fin shape follows the inverse square curve also to keep the greatest temperature differential for a longer time, which keeps the heat transfer rate higher for a longer time.

 

If you follow the patent, these shapes may be on either side or both (inside and outside) of an exchange system to speed up the rate of transfer of heat.

 

Particularly in steam systems, they promote both the formation of steam and on the condensation end they promote the condensation of the steam. By this speed up of reactions, an increase in the rate of energy transfer is obtained, and thereby reducing the amount of energy for heating or cooling.

Bob

[Mr Skeptic]

Right, the patent is about heat exchangers. I searched it for both "steam" and "partial" but could not find either. Your question here does not really depend on the heat exchanger, it's just assumed to be there.

 

Anyhow, if you want to test Dalton's law, you could change the partial pressure of a substance by heating it. You could add an inert gas (eg helium) and compare the difference. In fact, you can test it by just adding in a mix of various inert gases without any liquid.

[1bobwhite]

insane_alien,

Now to address what is really happening with air/steam mixtures, as I see it.

 

Once again bear with me as I wade through this explanation of why I say the interpretation of the data is incorrect.

 

Using the similar test arrangement, but instead of having pressure gauges, temperature probes are used, and instead of the gasses, steam is in chamber 1, and air is in chamber 2.

 

For the beginning of the test, the chamber pressures are at atmosphere.

 

Now at rest without steam, the temperatures are the same in both chambers.

 

When the steam valve is opened to chamber 1, 100 psi pure saturated steam at 328 degrees F. is introduced which expands the bag until the trapped air equalizes the pressure, at which point the steam ceases to enter, and the valve is closed.

 

Initially chamber 1 is at 328 deg.F, and chamber 2 is 68 deg. F. Progressively as latent heat is given up upon condensation of the steam on the bag wall, sensible heat is conducted through the bag wall so that the air is heating up as the steam is cooling. Since the steam was already at saturation, condensation procedes as the temperature cools, until all of the steam has condensed and the bag is collapsed.

 

The temperature readings at this point will be for chamber 1, the steam condensate temperature corresponding to the pressure that was developed by the heating and expanding of the air in chamber 2, but lower than the initial 328 deg F. Chamber 2 temperature will have increased until no more heat could transfer from the bag.

 

Now the heated air is in contact with the outer chamber wall so that it transfers its heat to the wall to be convected away on the outside of the wall.

 

Its temperature is lower than the initial input of 328 deg. F. by an amount depending how much air was involved.

 

This model represents what happens when air in the steam system "blanks out" and prevents the steam from transferring its latent heat directly to the vessel wall. As little as 5 % air can do this.

 

The next scenario is done without the bag.

 

Now without the bag, 100 psi pure saturated steam is introduced until flow stops, which means the intermixed air and the steam are at the same pressure.

 

The air temperature increased and the steam temperature decreased without condensation because of the warming airs ability to hold moisture. Condensation wont occur until both the air and the steam are at saturation for that temperature.

 

Now the rate of heat transfer through the vessel wall will depend on the sensible heat conducting through. The hot air and the steam molecules are both impinging upon the wall and releasing their heat.

 

When steam collides with a cooler surface and gives up its latent heat, its molecules collapse in volume and form a nearly instantaneous vacuum. This action could be thought of as popcorn poping in reverse. Going from a large kernal to a small seed. This vacuum then draws in more steam molecules that in turn collapse and give up their heat.

 

If the air is at say 30 % by volume then its molecules are hitting the wall 30% of the time, but blocking the steam molecule in the process for that 30 % of the time.

 

But air only gives up its sensible heat, and moves away, whereas steam when it contacts the wall gives up its sensible heat and its latent heat, but in the process it collapses by a significant amount thereby creating a "vacuum" that instantly draws in the next steam molecule.

 

Even though the temperature of both the air and the steam are the same within the vessel, the rate of heat transfer is reduced proportionally to the amount of air present because of this action of blocking of the steams condensation, with the result of a lower sensible heat at the outside of the exchanger.

 

See the image illustration of boundary layer action.

 

I hope this explanation helps.

Bob

[1bobwhite]

Mr Skeptic,

 

My patent is intended to be a generic patent that applies to any exchanger of energy, not just heat.

 

The heat exchanger shown is used as a generic example to make claims for the patenting process, even though it may be a useable design for some application.

 

Steam was not mentioned because it is specific to water, whereas the terms used are vapors, boiling, condensing, reheating, to describe actions of many volatile liquids that may be used in the device for purposes that dont necessarily involve heating and cool but for the condensation action. For instance the condensation of precious vapors of agents used in the perfume/fragrance industries.

 

Bob.

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insane_alien,

 

The topic of saturation is an involved one especially when steam and air mixes are concerned.

 

But in the steam tables I use, the definition is; Saturated Steam is pure steam at the temperature that corresponds to the boiling temperature of water at the existing pressure.

 

The tables are the numbers for pure saturated steam, not with any other contained gases.

 

The publication goes into the charts and graphs for descriptions for air in the system, how to handle the other gases, flash steam, etc, and the other aspects of steam that we need to know.

 

The objection I have with their explanations of heat exchangers, is in using the numbers from the steam tables for saturated steam to explain conditions that occur with steam that is not in saturation.

 

Also these pin finned heat exchangers described in my patent are directly related to the subject of steam, and air/steam mixtures in heat exchangers and how to improve their function.

Bob

[Skye]

 

Using the similar test arrangement, but instead of having pressure gauges, temperature probes are used, and instead of the gasses, steam is in chamber 1, and air is in chamber 2.

 

For the beginning of the test, the chamber pressures are at atmosphere.

 

Now at rest without steam, the temperatures are the same in both chambers.

 

When the steam valve is opened to chamber 1, 100 psi pure saturated steam at 328 degrees F. is introduced which expands the bag until the trapped air equalizes the pressure, at which point the steam ceases to enter, and the valve is closed.

 

Initially chamber 1 is at 328 deg.F, and chamber 2 is 68 deg. F.

So you begin with steam in chamber 1 at atmospheric pressure and 68 degrees F?

[insane_alien]

bob, you're going about testing daltons law all wrong by having a segregated system where daltons law SPECIFICALLY states it applies only in MIXED systems.

 

lets say you have a canister with 1 bar of nitrogen gas in side it.

 

to this canister you add carbon dioxide until the pressure is equal to 2 bar.

 

if you remove all the nitrogen leaving only the carbon dioxide the pressure in the canister will be 1 bar.

 

this means that both gases were producing 1 bar of pressure.

 

you cannot say that both gasses were at 2 bar because they were acting on the same surface. if the impacts from nitrogen alone were able to exhert a pressure of 2 bar and the molecules of carbon dioxide alone were able to eexhert a pressure of 2 bar then since both types of molecule are hitting the container wall with the same frequency and force then the system pressure MUST be 4 bar.

 

daltons law has been confirmed right down to molecular simulations of gases. to say it's wrong is actually crazy as the evidence is pretty irrefutable.

[1bobwhite]

insane_alien,

 

That is not true.

 

What my demo was showing is that it doesn't matter what element of gasses are involved, if they are mixed, it is impossible to tell experimentally which gas is giving which pressure.

 

If in your example, if you add one volume of gas, then of course it will increase the contained pressure, and conversely subtracting it will reduce that pressure, but you cant tell which one is causing the pressure when they are mixed.

 

In my demo, the first scenario had one gas separated by the bag membrane. It was not even coming into contact with the container wall to exert its pressure and yet it caused the contained pressure to increase as if it was present and impinging upon the wall.

 

So which gas is causing which pressure? Both chambers read the same pressure, which they would if they were mixed. You cant tell can you, since they are both providing pressure, yet one is not touching the wall.

 

It took up a volume that caused an increase in pressure. Its the gas volume that causes the pressure, and not its chemical element difference. If you add one mole of gas, it will increase pressure by that amount. Could care less if it is oxygen, nitrogen, whatever and you cant tell which one it is anyway.

 

I'll try it your way.

 

Lets say the container has a pressure of 100 psi.of air.

 

If the nitrogen is at 70% or a partial pressure of 70 psi, and oxygen is at 20% or a partial pressure of 20 psi, plus the rest at 10% or 10 psi, then these partial pressures add to 100 % or 100psi, correct? 70+20+10=100 psi

 

Refer to the attached illustration.

 

Now correct me if I'm wrong, the 70 psi nitrogen area would flow to the lower pressure areas until their pressures equalized, correct? And in doing so the nitrogen pressure would lower because of expansion, correct? So now the nitrogen is at a lower pressure, correct? The same for the oxygen at 20 psi, expanding into the other gas area of 10 psi, correct? What is the resultant overall pressure of the container? Is it still 100 psi?

 

How can you possibly have two or more different pressure areas within the same envelope without this equalization?

 

You cant' and that is the reason you cannot have partial pressures.

 

Ok, here is my challenge to you and anyone else that may read this thread.

 

For proving Daltons Law, show me the experimental apparatus and procedural data for this conformation and evidence you referred to, or where I may find this so that I may see for myself what you are talking about.

 

Not mathematical formulas and equations, but hardware and experimental procedure.

Bob

[Mr Skeptic]

What my demo was showing is that it doesn't matter what element of gasses are involved, if they are mixed, it is impossible to tell experimentally which gas is giving which pressure.

 

Which is also wrong. If this is what you're testing for, then you need a setup where it does matter which gas is giving which pressure. For example, get a mixture of He, H2O and CO2, with some water on the bottom. The amount of water that evaporates/condenses depends on the partial pressure of H2O. The amount of CO2 dissolved in the water depends on the partial pressure of CO2. The partial pressure of He will have nearly no effect whatsoever.

[1bobwhite]

Skye,

No, In this scenario, chamber 1 has the steam, chamber 2 has the ambient air at 68 deg F.

Regards, Bob

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Mr Skeptic,

 

For example, get a mixture of He, H2O and CO2, with some water on the bottom. The amount of water that evaporates/condenses depends on the partial pressure of H2O. The amount of CO2 dissolved in the water depends on the partial pressure of CO2. The partial pressure of He will have nearly no effect whatsoever.

 

In your example, several things are at play.

 

I'll use a two liter container of soda, if I don't drink it first, but it was topped off with helium to maintain the fizz.

 

The evaporation and condensation depend on both temperature and pressure. And total pressure, not just the supposed partial pressure of the water vapor.

 

The dissolved CO2 is a chemical association that is total pressure sensitive not just the supposed partial pressure of CO2.

 

And the helium boosted the total pressure to help keep the CO2 from disassociating from the water.

[insane_alien]

you are still treating these things as segregated systems. MIXED SYSTEMS CANNOT BE TREATED AS SEGREGATED SYSTEMS.

 

the nitrogen isn't going to flow, because it is already everywhere, the oxygen isn't going to flow because it is at the same partial pressure every where and so on.

 

really, this has been well studied and proven. if you wish to ignore facts then go somewhere else and do it because this is not the appropriate place.

 

seeing as you seem to be so intent on mis interpreting the soda bottle analogy, i'll give you another one.

 

you have a vessel filled with oxygen at 10kPa and another vessel containing oxygen with a partial pressure of 10kPa and nitrogen at 90kPa.

 

into these vessels you place a sample of lithium metal. which one will corrode the quickest?

 

using the current method, the samples will corrode at the same rate. using yours the sample in the nitogen/oxygen mix will corrode quickest.

 

as it happens, they corrode at the same rate. you can even try this yourself.

[1bobwhite]

insane_alien,

 

Please carefully consider the first experiment again.

 

For your stipulation, just remove the bag so that when your blow into the tube, the carbon dioxide freely mixes with the air that was in the container. You'll still end up with the container at the higher pressure without being able to demonstrate the partial pressure.

 

Lets say the gauges are located one on top and one on the bottom so that when CO2 is very slowly introduced, is stratifies on the bottom because of its greater density.

 

Now the CO2 is on the bottom, and air is on the top. The pressures will both read the same even though the smaller amount of CO2 is totally over the bottom gauge.

 

Now shake up the container so that the gasses mix, you'll still have the same total pressure on both gauges.

 

Next through one opening force in some soda water so that a pool of it is on the bottom, close off the valve, and notice the gauge readings are still equal even though one is submerged under soda water. Even as the soda water is further pressurizing the container, you will notice that the gauges will be equal throughout the whole exercise.

 

Now rotate the container so that the other gauge is submerged, and you'll still notice that the gauges are reading equal.

 

This experiment covers all of your objections, and can be done and verified by anyone.

 

Partial pressures CANNOT be experimentally demonstrated. And if you say they can, PROVE IT! By any source, by any repeatable test apparatus.

 

You say I'm am ignoring the facts. As I challenged you before, show me the testing equipment and test data from ANY source, especially the ones you are alluding to when you say

this has been well studied and proven.

 

And this is the place to explore and examine these so called facts you cite, and also when I carefully try to explain my position concerning the validity of this patent, which is the topic of this thread.

 

Respectfully, Bob

[insane_alien]

insane_alien,

 

Please carefully consider the first experiment again.

 

okay but you are still wrong.

 

For your stipulation, just remove the bag so that when your blow into the tube, the carbon dioxide freely mixes with the air that was in the container. You'll still end up with the container at the higher pressure without being able to demonstrate the partial pressure.

 

as the gas expands to equalise it's concentration in the container, the partial pressure will reduce. the system pressure however will remain identical.

 

Lets say the gauges are located one on top and one on the bottom so that when CO2 is very slowly introduced, is stratifies on the bottom because of its greater density.

 

Now the CO2 is on the bottom, and air is on the top. The pressures will both read the same even though the smaller amount of CO2 is totally over the bottom gauge.

 

first of all, they won't seperate that much. but seeing as they are separated, YOU DON'T HAVE A PERFECTLY MIXED SYSTEM WHERE DALTONS LAW APPLIES. YOU HAVE A (partially)SEGREGATED SYSTEM. how many times do i need to repeat myself here. mixed does not equal segregated, if you continue down this line i will be forced to assume you are being deliberately obtuse.

 

Next through one opening force in some soda water so that a pool of it is on the bottom, close off the valve, and notice the gauge readings are still equal even though one is submerged under soda water. Even as the soda water is further pressurizing the container, you will notice that the gauges will be equal throughout the whole exercise.

 

irrelevant

 

This experiment covers all of your objections, and can be done and verified by anyone.

 

no, it doesn't, you just think it does.

 

Partial pressures CANNOT be experimentally demonstrated. And if you say they can, PROVE IT! By any source, by any repeatable test apparatus.

 

perform any reaction that involves a gas phase. do it with and without an inert gas component. this will prove partial pressures as these will determine the rate of reaction. i have done these both theoretically and practically, you can find many papers on them in journals.

[1bobwhite]

insane_alien,

 

first of all, they won't seperate that much. but seeing as they are separated, YOU DON'T HAVE A PERFECTLY MIXED SYSTEM WHERE DALTONS LAW APPLIES. YOU HAVE A (partially)SEGREGATED SYSTEM. how many times do i need to repeat myself here. mixed does not equal segregated, if you continue down this line i will be forced to assume you are being deliberately obtuse.

 

You forgot to shake it.

Now shake up the container so that the gasses mix, you'll still have the same total pressure on both gauges.

 

Being obtuse? I have addressed every one of your objection directly with examples and diagrams and explanations.

 

The soda water was added to introduce another gas, water vapor, and dissolved CO2, to address one of the previous posted objections, that you are now saying is irrelevent.

 

as the gas expands to equalise it's concentration in the container, the partial pressure will reduce. the system pressure however will remain identical.

 

Now how can the partial pressure reduce and still have the overall pressure be the same? It can't. The total pressure will be less, by your example.

 

Your example of corrosion occurring under different gas pressures is introducing additional conditions that have different explanations for gas pressure changes.

 

How about this grade school experiment?

 

A burning candle is placed in the middle of a pan of water, and then a drinking glass is placed upside-down over the candle. The oxygen is consumed to CO2 creating a vacuum that draws the water up into the glass and extinguishes the candle.

 

Now the vacuum area above the water contains water vapor, CO2, air, and a little remaining oxygen.

 

Is this a suitable example showing consumption of oxygen in a container, comparable to your corrosion example? It is a little more dramatic.

 

Yet even in this vacuum, the partial pressures of the remaining gasses cannot be determined.

 

You keep referring to papers in journals and other sources to check out.

 

We are only a few keystrokes away from these sources on the internet, if you would be so kind as to find one so that I may read it and maybe I can be convinced that your points are valid.

 

Respectfully, Bob