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DTLarca
08-11-2010, 03:19 AM
Practical Implications of Dalton’s Law

Dalton’s law states that in a mixture of ideal gases, each gas has a partial pressure which is the pressure the gas would have if it alone occupied the volume.

The total pressure of a gas mixture is then the sum of the partial pressures of each individual gas in the mixture.

Because ideal gas molecules are so far apart they do not interfere with each other at all and so too each gas in a mixture behaves entirely as if it occupied the space alone.

What is Relative Humidity

Consider an absolute vacuum in a rigid cylinder which is equal in temperature to a 20°C ambient.

Water’s saturated vapour pressure at 20°C is 2340Pa. For any measured water vapour pressure there is only ever one associated dew point temperature. Here the dew point temperature is 20°C; dew point just is another name for saturated temperature, the line separating liquid from vapour.

Add pure water vapour to the cylinder to a pressure of 1170Pa. Since the water vapour is the only gas within the cylinder the total pressure in the cylinder would be 1170Pa and, by the way, the dew point temperature, or saturated temperature associated with 1170Pa, would be 9.27°C

The relative humidity (relative saturation) in the cylinder would be the ratio of the actual vapour pressure to the corresponding saturated vapour pressure for 20°C.

The saturated vapour pressure is that pressure that is theoretically possible according to the ambient temperature if sufficiently more water were added to the cylinder such that two more molecules added thereafter would also condense two molecules.

So relative humidity is:

Pactual/Psaturated = 1170/2340 = 0.5 or 50%RH

Maintained Independencelace of Partial Pressures

If atmospheric pressure is 101325Pa then to bring the total pressure within the cylinder up to atmospheric pressure we could add 100155Pa of dry air, which, with the existing water vapour pressure of 1170Pa, gives us the total of 101325Pa, all still at 20°C.

It is fundamental to the understanding of Dalton’s law of partial pressures that while the total pressure does now exceed the equivalent saturated vapour pressure for water at 20°C, being 2340Pa, the water vapour’s actual partial pressure remains at 1170Pa and so too the cylinder’s relative humidity remains at 50%. No condensation takes place, in fact, for reasons beyond the scope of this article; condensation is now even less likely.

If the water vapour pressure in the cylinder before adding dry air was instead a higher 2320Pa, meaning the dew point temperature was a higher 19.9°C, the relative humidity in the cylinder both before and after adding dry air would be 2320/2340 = 99%RH. Again, despite the relative humidity being so high no condensation would have occurred with any addition of dry air.

An Everyday Example Implication

When adding nitrogen to a system for pressure leak testing where a trace of the system’s refrigerant remains after recovery, the trace vapour will not condense regardless of how much nitrogen pressure is added to the system.

If the refrigerant in the system was R22 while the ambient temperature was still 20°C and the system charge was recovered until the remaining system pressure equalled atmospheric pressure then the R22 vapour relative humidity (relative saturation) remaining in the system would be:

Pact/Psat = 101325/909325 = 0.11 or 11%RH

Dalton’s law predicts that this relative humidity (relative saturation) will not change regardless of how much other gas you add to the system. Thus you cannot condense any amount of the trace gas with any amount of added nitrogen, so, the trace gas will continue to be of use for leak detection by methods seeking leaking refrigerant traces.

DTLarca
08-11-2010, 03:25 AM
Now that I can edit I decided to move what was here up to beneath what is above...

mad fridgie
08-11-2010, 03:49 AM
Very well written, and correct of course!
RH is very difficult for many to understand, hense the "sponge" Ok I understand you not your method.
What need to be brought into the sponge equation that the pressure reduces the sponge volume, thus changes the partial pressure of a fixed mass of water/refrigerant (A while since i was schooled "Boyles law"??)

DTLarca
08-11-2010, 04:20 AM
Very well written, and correct of course!
RH is very difficult for many to understand, hense the "sponge" Ok I understand you not your method.
What need to be brought into the sponge equation that the pressure reduces the sponge volume, thus changes the partial pressure of a fixed mass of water/refrigerant (A while since i was schooled "Boyles law"??)

Boyles is constant temperature.

Lussac's law is the relevant one here...

If you assume the cylinder volume is rigid thus it's volume does not reduce significantly while being cooled then the absolute pressure of the gas will reduce in proportion to the thermodynamic temperature.

Here's the tricky part - you cool the gas and its volume shrinks but the container's does not. So as it shrinks its pressure falls and so too then does its saturated temperature - the saturation temperature keeps falling away from the actual temperature each time you cool it some more - so what happens all along the way down to -273.15°C - does this tail chase continue all the way?

mad fridgie
08-11-2010, 07:44 AM
Boyles is constant temperature.

Lussac's law is the relevant one here...

If you assume the cylinder volume is rigid thus it's volume does not reduce significantly while being cooled then the absolute pressure of the gas will reduce in proportion to the thermodynamic temperature.

Here's the tricky part - you cool the gas and its volume shrinks but the container's does not. So as it shrinks its pressure falls and so too then does its saturated temperature - the saturation temperature keeps falling away from the actual temperature each time you cool it some more - so what happens all along the way down to -273.15°C - does this tail chase continue all the way?
Sorry for not explaining my self clearly.
It is the change in volume which is the point I was making. Of course should we then should account for the increase in temperature of the reducing vessel (shrinking sponge does not really apply)

DTLarca
08-11-2010, 09:44 PM
It is the change in volume which is the point I was making. Of course should we then should account for the increase in temperature of the reducing vessel (shrinking sponge does not really apply)

In the case of compression you should account for the increase in temperature of the reducing vessel - that is only when it decreases in volume on account of mechanical work. The only transfer of energy that does not require work is heat conduction.

But in the scenario I present above the vessel is assumed to remain rigidly constant in volume.

Different liquids and solids have their different coefficients of thermal expansion but all gases have the same coefficient of thermal expansion. Any gas at say 200K will shrink in volume by 1/200th of its initial volume when cooling from 200K to 199K. And any gas at say 100K will shrink in volume by 1/100th of its initial volume when cooling from 100K to 99K. And so on.

We can consider 3 different scenarios each where nitrogen, initially at 20°C, is exposed to liquid nitrogen -196.8°C

Nitrogen gas in a balloon and the balloon under atmospheric conditions.
Nitrogen gas of the atmosphere
Nitrogen gas in a rigid container.
What % diameter does the balloon reduce to?
What happens to the saturation point in the second scenario?
When does the nitrogen condense in the third scenario?

mad fridgie
08-11-2010, 10:17 PM
1, Volume would reduce to 1/216.8, so diameter would roughly by 84%
2 remins the same unless the atmosphere componets condense, then Nitrogen proportion changes, then saturation would change
3 It does not, because the pressure would reduce

DTLarca
08-11-2010, 10:41 PM
1, Volume would reduce to 1/216.8, so diameter would roughly by 84%
2 remins the same unless the atmosphere componets condense, then Nitrogen proportion changes, then saturation would change
3 It does not, because the pressure would reduce

All good on scenarios 2 and 3. I gave scenario 1 first to try make you think it might be the easiest - it's the tricky one :)

The volume should shrink to 76/293 of its original which is 26% of its original. But by how much do you reduce the diameter to for a 74% reduction in volume? Volume is proportional to diameter cubed so the cubed root of 26 must be the new percentage volume - the cubed root of 26 is about 64. But in the fact when you dip the balloon into liquid nitrogen nitrogen liquid evaporates taking with it the heat of the nitrogen gas in the balloon thereby condensing the nitrogen gas in the balloon causing the balloon to collapse completely containing a minute amount of liquid nitrogen :)

And there starts the journey through the fundamentals of thermodynamics and the branch we call psychrometrics :)

DTLarca
08-11-2010, 10:49 PM
The volume should shrink to 76/293 of its original which is 26% of its original. But by how much do you reduce the diameter to for a 74% reduction in volume? Volume is proportional to diameter cubed so the cubed root of 26 must be the new percentage volume - the cubed root of 26 is about 64.

I'm being lazy here - typing out as if I was talking it out - sort of thing.

I shouldn't be saying 26 and 64 but rather factors 0.26 and 0.64. The cubed root of 0.26 is about 0.64 but the cubed root of 26 is not 64 :)

mad fridgie
08-11-2010, 10:51 PM
All good on scenarios 2 and 3. I gave scenario 1 first to try make you think it might be the easiest - it's the tricky one :)

The volume should shrink to 76/293 of its original which is 26% of its original. But by how much do you reduce the diameter to for a 74% reduction in volume? Volume is proportional to diameter cubed so the cubed root of 26 must be the new percentage volume - the cubed root of 26 is about 64. But in the fact when you dip the balloon into liquid nitrogen nitrogen liquid evaporates taking with it the heat of the nitrogen gas in the balloon thereby condensing the nitrogen gas in the balloon causing the balloon to collapse completely containing a minute amount of liquid nitrogen :)

And there starts the journey through the fundamentals of thermodynamics and the branch we call psychrometrics :)
Good trick question, can not argue with that one!

mad fridgie
08-11-2010, 11:01 PM
We have two non dimensional containers (sealed)
One containing a % by mass of vapour at saturation point at one pressure, and the second containing the same (working fluid) with a different % by mass at saturation point at different pressure. If we had a valved pipe between the 2 (which for argument has no voulme) How would you calculate the equalibrium pressure and temperature (NO heat gains and losses)

DTLarca
08-11-2010, 11:07 PM
We have two non dimensional containers (sealed)
One containing a % by mass of vapour at saturation point at one pressure, and the second containing the same (working fluid) with a different % by mass at saturation point at different pressure. If we had a valved pipe between the 2 (which for argument has no voulme) How would you calculate the equalibrium pressure and temperature (NO heat gains and losses)

% mass of the total mass of each sample multipled by each mass it's own absolute pressure and it own absolute temperature.

DTLarca
08-11-2010, 11:09 PM
If the mixed temperature is lower than the mixed saturation temperature based on the mixed pressure then there will be condensation.

mad fridgie
08-11-2010, 11:17 PM
% mass of the total mass of each sample multipled by each mass it's own absolute pressure and it own absolute temperature.
Could you use enthalpy?

DTLarca
08-11-2010, 11:27 PM
But I am using enthalpy - just not explicitly :)

PV=nRT where because we are using moles (n) not mass (m) R therefore is the universal gas constant which is kJ/kg (enthalpy) per K.

Then by using % mass I am short cutting past using mole counts and molar masses.

Make sense?

mad fridgie
08-11-2010, 11:28 PM
% mass of the total mass of each sample multipled by each mass it's own absolute pressure and it own absolute temperature.
And! We now have a total value, neither a temperature or pressure

DTLarca
08-11-2010, 11:30 PM
And! We now have a total value, neither a temperature or pressure

My off the cuff reply here has to be to ask if you understand Avogadro's principle - I cannot see how you could validate that claim?

mad fridgie
08-11-2010, 11:40 PM
My off the cuff reply here has to be to ask if you understand Avogadro's principle - I cannot see how you could validate that claim?
As a dyslectic, I do not remember names but generally principles. So no I do not Avogadro

So we (% * AbP * AbT) + (% * AbP* Abt) = *****
correct so far

% of 100%

DTLarca
08-11-2010, 11:43 PM
And! We now have a total value, neither a temperature or pressure

If the mixture temperature is below the saturation temperature for the pressure - on instantaneously mixing - then there will be fogging and then condensation of the excess vapour beyond what the enthalpy is capable of supporting.

Okay - so now I have woken up to your insistence on knowing enthalpy :)

How then would I determine the enthalpy which then in turn would help me determine what quantity by mass of the mixture condenses on mixing to a mixed temperature below the mixed saturated temperature... back soon...

DTLarca
08-11-2010, 11:55 PM
Going from moles to mass therefore from universal to specific gas constants...

We have R for each gas:
(PV)/(mT) = R

We have the mass, pressure and temperature of each gas but not each gases volume.

So we determine each volume:
V = (mRT)/P

Knowing each volume we can determine the enthalpy kJ/kg by multiplying each R (kJ/kg.K) by each samples absolute temperature which cancels to give kJ/kg (kJ/kg.K x K)

Like this...

PV/m = RT where RT is enthalpy.

So knowing P, V and m for each sample we know RT (Enthalpy) for each sample and so the mixed enthalpy is the total enthalpy and from that we can look up on a PH chart our percentage quality :)

How's that look?

mad fridgie
08-11-2010, 11:59 PM
The more I know, the more I know, I know nothing!
When dealing in peer groups, it is normally presumed that the levels knowledge within the peer group are similar, in generally termonlogy that is true, but specifically/detailed this less than the case, do you nod your head, and let the specfics go by, or do act a little bit stupid (more stupid than you are on a particular subject) to ensure that the expert spends sufficient time to explain there levels of expertise, so that the detail can be retained by the more stupid individual.
I choose to be the stupid one.
I employ this method when learning or teaching.
So when teaching I do use "this statement is not quite technically true" if the statement is better suited for the indiviual to grasp a base principle. Over time true techincal aspects can be reinforced.

mad fridgie
09-11-2010, 12:13 AM
Going from moles to mass therefore from universal to specific gas constants...

We have R for each gas:
(PV)/(mT) = R

We have the mass, pressure and temperature of each gas but not each gases volume.

So we determine each volume:
V = (mRT)/P

Knowing each volume we can determine the enthalpy kJ/kg by multiplying each R (kJ/kg.K) by each samples absolute temperature which cancels to give kJ/kg (kJ/kg.K x K)

Like this...

PV/m = RT where RT is enthalpy.

So knowing P, V and m for each sample we know RT (Enthalpy) for each sample and so the mixed enthalpy is the total enthalpy and from that we can look up on a PH chart our percentage quality :)

How's that look?
It has been knocking 30 years, since schooling, and you tend not use the pure knowledge gained, but data that relates directly with the industry, and terms that go with the industry. (even though in many cases it is the same thing)
We have just completed in the last few minutes, a scientific ideal, into a method of which those in the refrigeration could use (even though in this case, not that much use for the everyday project)
Thank you

DTLarca
09-11-2010, 12:24 AM
The more I know, the more I know, I know nothing!

Absolutely everything is energy. There are only two types of energy - kinetic or potential.

Even E=mC² is just a transformation of KE = 1/2 m v²

And in any system the energy can either be all kinetic or all potential such as a roller coaster up high but slow or way down and fast. Thus:

½mv² = mgh

rearranging the equating of KE and PE enables you to determine everything and understand or interpret or describe everything.

Even the time for a body to fall from a height you expand to reveal time and then rearrange:

½mg²t² = mgh
t = √(2h/g)

Everything from the speed of a bullet to stack effect or fans or orifices and then onto fluid dynamics then acoustics - well - everything is then understood.

One of my HND students - who frequents this board - told me he wished he could go back and start his HNC and HND all over again after seeing the equating of KE and PE used to illustrate a bunch of the other areas of engineering like fans and pumps and so on.

DTLarca
09-11-2010, 12:31 AM
It has been knocking 30 years, since schooling, and you tend not use the pure knowledge gained, but data that relates directly with the industry, and terms that go with the industry. (even though in many cases it is the same thing)
We have just completed in the last few minutes, a scientific ideal, into a method of which those in the refrigeration could use (even though in this case, not that much use for the everyday project)
Thank you

Cool - that was a fun question to answer - I have not been given a question like that since I was an apprentice. I cracked all of the puzzles given to me like this one back in those days.

Actually - I was given a few similar question by a guy in the UK called Mike Creamer at the pub on two seperate occasions - I cracked those puzzles too - so now everytime we run into each other he tells me I'm the smartest refrigeration dude he has ever met - next time I will get it in writting - I just love this **** - that's all :)

mad fridgie
09-11-2010, 12:33 AM
It was a reference to life in general!
I to wish I had made better use of possible education, but chasing girls and earning a quid was so more important! You can not put a old head on young shoulders. (well on these shoulders anyway)
I also agree everything is energy (although the only two princples I can not grasp are the big bang theory and god)

DTLarca
09-11-2010, 12:48 AM
It was a reference to life in general!
I to wish I had made better use of possible education, but chasing girls and earning a quid was so more important! You can not put a old head on young shoulders. (well on these shoulders anyway)
I also agree everything is energy (although the only two princples I can not grasp are the big bang theory and god)

Here is a great video - also asserts KE and PE as the only known energy forms in the Universe.
http://www.youtube.com/watch?v=FUfXJ1Qszgk

Then this is probably the best speech available on the origins of the universe - how the laws of physics point at the universe most likely having come from nothing.
http://www.youtube.com/watch?v=7ImvlS8PLIo

DTLarca
09-11-2010, 01:05 AM
It was a reference to life in general!
I to wish I had made better use of possible education.

I don't stop - you should just keep going - I'm reading this sort of stuff all the time. My personal library has about £3000.00 worth of books in it - mostly philosophy but a fare deal of physics too.

mad fridgie
09-11-2010, 01:18 AM
I don't stop - you should just keep going - I'm reading this sort of stuff all the time. My personal library has about £3000.00 worth of books in it - mostly philosophy but a fare deal of physics too.
Unfortunately I can not read (more than just a few words), reading talks a massive effort (and very tiring). this is something that is difficult to understand. The longer the the writing, the longer it takes to digest. I have never had the pleasure of reading a fiction book, even though I have tried so many times. Reading a words is not normally a problem, understanding the words is!

Gary
09-11-2010, 01:44 AM
It's interesting how different people learn.

I find that lectures are wasted on me. The spoken words just kinda blend together and the mixture is meaningless.

Back in school, I wrote copius notes and learned from writing and reading the notes.

Different strokes for different folks. :)

DTLarca
09-11-2010, 07:45 AM
It's interesting how different people learn.

I find that lectures are wasted on me. The spoken words just kinda blend together and the mixture is meaningless.

Back in school, I wrote copius notes and learned from writing and reading the notes.

Different strokes for different folks. :)

Socrates and Plato insisted on groups in discussion no larger than 5.

I am similar, I think, Gary. I have to read each single new principle, from a text book, then go and play with it on paper. Then come back and read the next 1/4 or 1/3 of the page before going back to the spreadsheet or paper etc.

Then a week later I forget. So I have to do it all over again - but this time quicker. After the 3rd time through the stuff it might start sinking in for a permanent run. But its the kick in the discovery of a new angle or a whole new concept that is addictive.

I tend not to write articles about stuff I know well - I write about stuff I want to know well. Since I've done a bit of lecturing I now make power point presentation of stuff I am learning or even of any system I am designing. Then I can present it to others whenever the occasion arises - and they do - and I get to know the stuff a whole lot better - like really a whole lot better.

I see physics as an abstract Rubik's cube to play with though it is ultimately empirical in nature MKSΘ - philosophy is for the going where physics can't - if it can't be answered with the aid of a ruler, weigh scale, stop watch or thermometer then it requires philosophising.