Boiling off Liquid Argon Using A High Powered Resistor

  • #1
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I understand that liquid argon is constantly boiling when it's in contact with the air in the lab, but I want to increase that boil off rate. Because of that, I got myself a high powered resistor (1 ohm, 100 W) and got the highest current power supply I had (5A). When I supply about 5 V and 5 A to it and leave it in air, it heats up quite a lot. I'm not exactly sure what the exact temperature is after like 3 minutes of running it, but if I put ice on it, it instantly melts it and then boils off the water.

I then put this resistor inside a dewar and fill it up with liquid argon and measure the amount of time it takes for the volume to boil off without the resistor being powered on. It takes about an hour. I then fill back in the same volume of argon and then supply current to the resistor and record the time and it comes out to be about an hour. It seems to me that having the resistor made no difference. In addition, when I take out the resistor (while it's still powered on and it's immersed in liquid), it's cold as ice! After I wait for a few minutes and leave it in air, it heats back up again. My guess for why this is happening is that there's not enough heat that the resistor is able to supply to boil off the liquid. Is that so? Will having multiple resistors then and a larger current source (so I can have a total power dissipation of, say 500W) solve the problem? Has anyone done this before?

I do have a cheaper solution, as opposed to buying an expensive high current power supply, which is to just surround my dewar with heating tape and crank it up to about 80-100 deg C, but I am really curious as to why my current method doesn't work. By the way, I wrote liquid argon everywhere since that's what we primarily use to do our experiments, but I did this specific test using liquid nitrogen. I suspect it would be even easier to increase the rate of boiling of nitrogen than argon since nitrogen's boiling point is lower than that of argon.

Thanks!
 

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  • #2
mjc123
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How big is your dewar? A rough calculation suggests your resistor would boil off ca. 500 mL liquid Ar per hour. Is that significant compared to the volume of your dewar? But how good is your dewar anyway, if it boils off in an hour?
 
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  • #3
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A better power supply and more resistors will help. If you have a lot of liquid argon or nitrogen then 25 W won't do much (you should be able to calculate that). To make it worse, the resistance of the resistor will probably be lower at cryogenic temperatures. Cable losses could be a concern, too.

If you have some place with a large surface area then you could spread out the liquid there and let the environment boil it off.

Make sure the place is well ventilated if this is indoors.
 
  • #4
256bits
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Summary: How do I boil off Liquid Argon using a high powered resistor (100W), is it at all possible?

It takes about an hour
You need a better clock. Maybe one that counts minutes and seconds, rather than "abouts".
 
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  • #5
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I suspect it would be even easier to increase the rate of boiling of nitrogen than argon since nitrogen's boiling point is lower than that of argon.
It’s not the boiling point you care about it, it’s the latent heat of evaporation (the energy required to turn a liquid at the boiling point into vapor at the boiling point) that matters here. Google and a back of the envelope calculation will tell you how much heat you have to add to the dewar; once you know that you can consider how to produce that heat and whether your power supply is up to the job.
 
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  • #6
Vanadium 50
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It’s not the boiling point you care about it, it’s the latent heat of evaporation (the energy required to turn a liquid at the boiling point into vapor at the boiling point) that matters here.

This.

Also, the fact that nitrogen boils at 77K and argon at 88K is irrelevant because your liquid is at its boiling point, whatever it is.

Getting back to the problem at hand, 5V at 5A is 25W. The fact that the resistor is rated for 100W is irrelevant. The energy supplied to the dewar is therefore 90 kJ. Spending a few minutes with Google is a good idea, but half a liter per hour doesn't sound crazy. This is the equivalent to 3 drops per second.

More things to think about. Where is your resistor located. How do you know it is 1 ohm? More importantly, how do you know it is 1 ohm at 77K? Is your power supply constant voltage or constant current?
 
  • #7
Phys12
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How big is your dewar? A rough calculation suggests your resistor would boil off ca. 500 mL liquid Ar per hour. Is that significant compared to the volume of your dewar? But how good is your dewar anyway, if it boils off in an hour?
So, I'd say I have about 10-20 L of liquid nitrogen. Regarding how good the dewar is, the one I'm using isn't the best and we typically surround our dewar with liquid argon, I didn't do this time since it was just a test.
 
  • #8
Phys12
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You need a better clock. Maybe one that counts minutes and seconds, rather than "abouts".
Hahaha! I did use a stopwatch which can calculate up to seconds, but since it took 59 minutes, I didn't think it was too significant to mention that and just say about an hour. And what I really care about is the difference in boil off rate, not the absolute rate and I could figure that out with such approximation (I didn't mention 59 minutes in the question, but that's what I measured in both cases).
 
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  • #9
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This.

Also, the fact that nitrogen boils at 77K and argon at 88K is irrelevant because your liquid is at its boiling point, whatever it is.
I agree that the temperature is irrelevant. But it does matter that the two liquids are different, right? Given that their heat capacities will be different because of this equation: Q = mCΔT (https://www.kentchemistry.com/links/Energy/SpecificHeat.htm)

Getting back to the problem at hand, 5V at 5A is 25W. The fact that the resistor is rated for 100W is irrelevant. The energy supplied to the dewar is therefore 90 kJ. Spending a few minutes with Google is a good idea, but half a liter per hour doesn't sound crazy. This is the equivalent to 3 drops per second.

More things to think about. Where is your resistor located. How do you know it is 1 ohm? More importantly, how do you know it is 1 ohm at 77K? Is your power supply constant voltage or constant current?
I agree that the rating of the resistor is irrelevant as long as I'm dissipating less power than the maximum that the resistor is capable of.

So, my dewar eventually will be closed (I left it open for the test since I cared more about the *difference* in boil off rate). The resistor is put at the bottom of the dewar. I did do a back of the envelope calculation and the energy I got was like 3600 kJ of energy... but after talking to a grad student, we thought that *any* heat would probably be enough to increase the rate of boiling by a lot since liquid argon boils even at room temperature. But after doing the experiment, I trust the calculations more and believe that we really need hundreds, if not thousands, of watts to get a sufficient boil off rate.

To answer your questions, I know it's 1 ohm because that's what they're rated to be and I measured using a multimeter. I do not think they're 1 ohm at 77K, but I thought it's irrelevant what the resistance is as long as I know what the power dissipated is, which I could calculate given what the current drawn from the power supply was and the voltage output. The way I work with my power supply is I either set the voltage to a specific value and if it's low enough, that limits the current drawn or I set a specific current, while having a high limit on the voltage and that limits the amount of voltage drawn. My power supply can give out max. of 30 V and 5 A so I set the voltage to be 10V and current to be 4.9A and it draws 4.9A of current and the voltage is at about 5V.
 
  • #10
russ_watters
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I agree that the temperature is irrelevant. But it does matter that the two liquids are different, right? Given that their heat capacities will be different because of this equation: Q = mCΔT (https://www.kentchemistry.com/links/Energy/SpecificHeat.htm)
No, that equation describes the amount of energy required to raise the temperature of the liquid (or vice versa). But your liquid temperature is fixed, right? So it doesn't apply, right?

Again: what matters is the power output of the resistor relative to the heat of vaporization of the liquid.
So, my dewar eventually will be closed (I left it open for the test since I cared more about the *difference* in boil off rate). The resistor is put at the bottom of the dewar. I did do a back of the envelope calculation and the energy I got was like 3600 kJ of energy...
In what time period? ...I'm pretty sure that's wrong. Please show the math you used.
...but after talking to a grad student, we thought that *any* heat would probably be enough to increase the rate of boiling by a lot since liquid argon boils even at room temperature.
That makes no sense at all - it doesn't reflect an understanding of the thermodynamics of this situation at all, so I hope you misunderstood this grad student. This is a really straightforward heat vs mass transfer problem. Freshmen physics or chemistry classes often include a lab where they put a pot of water on a hot plate and measure/calculate how long it will take to boil off. Same exact problem. Please try to do the math here, as we've described the problem; and if you are still unsure, ask where you are having an issue with it.
 
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  • #11
Vanadium 50
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Please try to do the math here, as we've described the problem; and if you are still unsure, ask where you are having an issue with it.

One number that will be illuminating is the total energy needed to boil the LN2 away (let's not talk about argon, because from the description argon is not involved).
 
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  • #12
russ_watters
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One number that will be illuminating is the total energy needed to boil the LN2 away (let's not talk about argon, because from the description argon is not involved).
Yes, and then divide that by the power output of the resistor...
 
  • #13
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To answer your questions, I know it's 1 ohm because that's what they're rated to be and I measured using a multimeter. I do not think they're 1 ohm at 77K, but I thought it's irrelevant what the resistance is as long as I know what the power dissipated is
Some of that power will be dissipated in cables instead of the resistor.
So, my dewar eventually will be closed (I left it open for the test since I cared more about the *difference* in boil off rate).
Why would you close it if you want to get rid of the liquid as quickly as possible?
In what time period? ...I'm pretty sure that's wrong. Please show the math you used.
It's an energy, it doesn't have a time period, and the number looks plausible.
That makes no sense at all - it doesn't reflect an understanding of the thermodynamics of this situation at all, so I hope you misunderstood this grad student.
Huh? Quite sure OP meant room temperature for the environment, which does indeed make nitrogen or argon boil, and adding a heater will speed up this process.
 
  • #14
russ_watters
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It's an energy, it doesn't have a time period...
It does if he's calculating the amount of energy added to the dwar by his resistor. The entire point of this exercise is to decrease time.
...and the number looks plausible.
My read of the statement was that he's saying the energy added by the resistor is 3,600 kJ. V50 is right that it is 90 kJ; 25 watts for an hour. If the OP was thinking of his resistor being a 100W resistor, he's still off by a factor of 10.

Perhaps you are thinking he's saying that's the heat of vaporization. That could be - 3,600 kJ would be close, but I couldn't duplicate it. In context it doesn't look like it to me. The OP can clarify.
Huh? Quite sure OP meant room temperature for the environment, which does indeed make nitrogen or argon boil, and adding a heater will speed up this process.
Here's that sentence again:
...but after talking to a grad student, we thought that *any* heat would probably be enough to increase the rate of boiling by a lot since liquid argon boils even at room temperature.
My read is that the grad student thought adding even a tiny heater to the dewar would vastly increase the boiling rate. That's why the OP is here -- he put in a tiny heater and it didn't increase the boiling rate much.

Also, the fact that the dewar is sitting in room temperature air doesn't have anything to do with it, so that part is an odd thing to add...unless it is connected to the OP's confusion about why the resistor was cold when put into the dewar. It strikes me as a confusion about the difference between temperature and heat. But I'm less sure about that.
 
  • #15
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My read of the statement was that he's saying the energy added by the resistor is 3,600 kJ.
I read the 3600 kJ as energy needed to boil away the nitrogen. That is in the right range.
My read is that the grad student thought adding even a tiny heater to the dewar would vastly increase the boiling rate.
OP didn't specify the rate there. Just "faster". Which is qualitatively correct.
 
  • #16
Vanadium 50
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The whole "faster" discussion for some reason reminds me of the old "you're in a balloon" joke.

The OP was surprised that his heater didn't make a substantial difference. Comparing the energy it takes to boil the LN2 with the energy supplied would have cleared this up: he's adding a few percent to the existing heat leak. (Aside: he has a heat leak of order a kilowatt. Yowza!)

The OP also has a misunderstanding of latent heat and heat capacity, and likely one about temperature and heat as well. It would be good to clear this up because a) it's what we do at PF, b) he's working with cryogens and it is therefore a safety issue, and c) he has ambitions to go to elite graduate schools, and they will expect him to have learned this.
 
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  • #17
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No, that equation describes the amount of energy required to raise the temperature of the liquid (or vice versa). But your liquid temperature is fixed, right? So it doesn't apply, right?
Yeah, I think that's probably right. I say probably because I think it's applicable to this specific test that I did, but maybe not in the final experiment. The reason why I used the equation of heat capacity is because this is what my actual experiment was going to look like: I would first put my dewar in high vacuum and then fill it with liquid argon. After that, I made a crude approximation to find out the power I may need and assumed that initially the argon wouldn't be boiling at all (since it's not at room temperature as the dewar was closed and evacuated) and I would have to raise the temperature of it in order to make it boil. I didn't go through the process of first calculating the heat it takes to raise the liquid to its boiling point and then using specific heat to figure out how much of the liquid I'm converting to gas in what amount of time given the amount of heat I'm adding. That was my bad

Again: what matters is the power output of the resistor relative to the heat of vaporization of the liquid.
That's what I thought too and that's why I did the calculation first to find out exactly what power I needed.

In what time period? ...I'm pretty sure that's wrong. Please show the math you used.
So, I think that calculation is wrong, but for a different reason. Reason being that I calculated only the heat needed to raise the temperature of the liquid by a few Kelvin, but not the heat required to convert that volume to gas (note that the calculation which yielded 3600 kJ was to raise the temperature of about 100 L of liquid argon to about 10K). I used 10K since the difference in the temperature of liquid argon and its boiling point is about a few Kelvin, I think the time period would then be one hour.

Some of that power will be dissipated in cables instead of the resistor.
Will it be significant? Because when I have the resistor our in air, it get pretty warm (about 80 deg C or so) so I thought it's probably fine to ignore the dissipation from cables.

Why would you close it if you want to get rid of the liquid as quickly as possible?
Because the experiment will be done in high vacuum conditions. We first evacuate our cryostat and then fill it with liquid argon and do whatever experiments we need to do. In order to achieve high vacuum, we need to close the dewar...

My read of the statement was that he's saying the energy added by the resistor is 3,600 kJ
No no, maybe I didn't put my words clearly. That's the energy I calculated that I *need*. My power output with the current resistor is only 25 W, which will give me an energy of 90 kJ in an hour.

Also, the fact that the dewar is sitting in room temperature air doesn't have anything to do with it, so that part is an odd thing to add...
I don't think it does either since I'm looking at the difference in boil off rate when I have the resistor powered on vs. when it is not. I thought to mention it to fully describe my setup

The OP also has a misunderstanding of latent heat and heat capacity, and likely one about temperature and heat as well.
I don't think I have a misunderstanding of either. I simply wasn't thorough enough in my calculations, which was my mistake.
 
  • #18
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My read is that the grad student thought adding even a tiny heater to the dewar would vastly increase the boiling rate. That's why the OP is here -- he put in a tiny heater and it didn't increase the boiling rate much.
Correct. I don't think the grad student thought it through, maybe he was in a hurry?
 
  • #19
DaveC426913
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If you have some place with a large surface area then you could spread out the liquid there and let the environment boil it off.
This should help a lot. Maximizing surface area should greatly increase evaporation rate.

* see sig-line caveat
 
  • #20
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I don't think I have a misunderstanding of either. I simply wasn't thorough enough in my calculations, which was my mistake.

Given that their heat capacities will be different because of this equation: Q = mCΔT

No matter how thorough you are with Q = mCΔT you will not get the right answer. Because...

It’s not the boiling point you care about it, it’s the latent heat of evaporation

Again: what matters is the power output of the resistor relative to the heat of vaporization of the liquid.

If you are insisting that the wrong equation will give you the right answer, yes, you have a misunderstanding.
 
  • #21
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This should help a lot. Maximizing surface area should greatly increase evaporation rate.

* see sig-line caveat
My dewar is of a constant volume, I cannot change that during the experiment. I could transfer it to a different dewar and do the experiment there but because the volume will be greater in it, along with the surface area, I don't think it'll make a difference

No matter how thorough you are with Q = mCΔT you will not get the right answer. Because...
Correct, but if I wrote it right above, and maybe I didn't convey it properly, I said that my lack of thoroughness came from the fact that I didn't include latent heat in my calculation to actually account for converting the liquid to gas. I only considered the part of raising the liquid's temperature to its boiling point.

If you are insisting that the wrong equation will give you the right answer, yes, you have a misunderstanding.
I'm not insisting that the wrong equation will give me right answer. To the contrary, I admitted that my first calculations would've given me an incorrect answer. To be precise, it would have giving me lower heat than I actually need since I first need to heat the liquid to its boiling point (which I calculated) and then convert it to gas (which I missed). I don't understand where I insisted that the wrong equation will give me the right answer...I may have insisted that I do not have a misunderstanding when it comes to knowing the difference between heat and temperature and between latent heat and heat capacity, which I think I'm right about.
 
  • #22
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Maximizing surface area should greatly increase evaporation rate.
I do not agree that evaporation rate is the crucial factor here. It is the rate at which heat is added to the fluid. As long as the fluid is at boiling temperature, evaporation will take place at whatever rate is needed to balance the heat flow.

Increasing surface area will increase heat flow which will, in turn, increase the evaporation rate.
 
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  • #23
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I would first put my dewar in high vacuum and then fill it with liquid argon. After that, I made a crude approximation to find out the power I may need and assumed that initially the argon wouldn't be boiling at all (since it's not at room temperature as the dewar was closed and evacuated) and I would have to raise the temperature of it in order to make it boil.
I don't understand this.
You are putting liquid argon into an evacuated chamber and then attempting to heat the liquid?
If so, then you need to definitely take into account the vapour pressure as the temperature of the liquid rises.
We sure do not want your sealed chamber experiencing a too too much pressure stress.
Does it have a relief valve to limit the maximum pressure?

You have to look at the vapour pressure vs temperature of argon and do some more pretty fine calculations before you preform this experiment.

I couldn't find anything for argon.
But nitrogen.
http://www.ddbst.com/en/EED/PCP/VAP_C1056.phpAt atmospheric pressure, liquid nitrogen boils at 77K.
If you increase the temperature of the liquid, which you would be doing in the closed chamber, the vapour pressure also increases. Cryogenic liquids have an expansion of around 700.
if you look at the chart, 101 kPa corresponds to a boiling temperature of 77K
A rising temperature, ( the chart does not go to room temperature ) - at 84 K you are already at 2 atm.
at 87 close to 3 atm
at 91 close to 4 atm.
and so on

No wonder that containers transporting cryogenic liquids, or containing them, are rated as being able to do so.
 
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  • #24
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Does it have a relief valve to limit the maximum pressure?
Yes
 
  • #25
256bits
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By the way, if I understand the setup correctly, I don't think you will see any boiling with a minimal heat addition, just a slow rise in temperature.
 
  • #26
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Will it be significant? Because when I have the resistor our in air, it get pretty warm (about 80 deg C or so) so I thought it's probably fine to ignore the dissipation from cables.
See above, the resistance will probably drop with temperature. You can measure it. Cool it, then take it out and quickly measure the resistance with short cables, or (better) do a 4-point measurement with the resistor in the liquid. You can even use the heating setup for this measurement: Add two cables to measure the voltage drop at the resistor, you already know the current.

If this will be in a vacuum later then heat transfer from the environment to the dewar will be greatly reduced. You should take this into account.

Why is it so important to get rid of the liquid nitrogen or argon remotely without access to it?
 
  • #27
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See above, the resistance will probably drop with temperature. You can measure it. Cool it, then take it out and quickly measure the resistance with short cables, or (better) do a 4-point measurement with the resistor in the liquid. You can even use the heating setup for this measurement: Add two cables to measure the voltage drop at the resistor, you already know the current.

If this will be in a vacuum later then heat transfer from the environment to the dewar will be greatly reduced. You should take this into account.
Ok, that sounds good, I'll try that. Thank you

Why is it so important to get rid of the liquid nitrogen or argon remotely without access to it?
I'm trying to recycle liquid argon in order to not waste it. To do so, I wanted to pass the gas from the dewar through a bunch of filter materials to increase its purity (since I expect the purity of liquid argon to decrease over time and I want to keep that constant). For that, the setup I had in mind was to pass the gas that's boiling off the liquid argon through some filter materials to remove Oxygen and Water (impurities) from it and then pass it through a liquid argon bath to condense it back to liquid argon and then have that enter back into the dewar. In order to get a high enough gas pressure to be able to push the liquefied argon back into the dewar, I need a high boil off rate of argon.
 
  • #28
cjl
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If the whole goal is to minimize waste of argon, this seems like a very strange way to do it. You're likely better off trying to minimize your evaporation rate in the first place rather than trying to recirculate it like this.

In addition, if you're boiling off a mixture of oxygen, water, and argon, you likely won't get much oxygen or water in the vapor. They'll stay in the liquid, and your vapor will be almost entirely argon, since it's the component with the lowest boiling point (though water can't stay in the liquid at any significant concentration anyways, due to being so far below the freezing point, though I admit I don't know the exact miscibility of water and liquid argon and what it does to the phase diagram). This is fundamentally how distillation works.

Finally, if you do get this working, how do you expect to keep your condensing argon bath cool? If you're using it to condense argon vapor in your return path, you'll have just as much evaporation from the bath as you have condensation in the tube, which isn't great for your low waste goals.
 
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  • #29
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If the whole goal is to minimize waste of argon, this seems like a very strange way to do it. You're likely better off trying to minimize your evaporation rate in the first place rather than trying to recirculate it like this.
The goal is to recycle argon, while also increasing its purity. Over time, we expect the purity of argon to decrease and we're hoping this to help with that.

Finally, if you do get this working, how do you expect to keep your condensing argon bath cool? If you're using it to condense argon vapor in your return path, you'll have just as much evaporation from the bath as you have condensation in the tube, which isn't great for your low waste goals.
Yeah, that's a good question and something I've been trying to figure out myself...
 
  • #30
russ_watters
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Over time, we expect the purity of argon to decrease...
Why? That seems very unlikely to me. Moreover, a kludged moonshine distilling setup is probably more likely to add impurities to the argon than remove them!

Do you have any chemists/chemical engineers advising on this work? What you are doing is very odd...

What is the actual experiment you are trying to perform and what is required of it in terms of atmosphere? For what course and level is this project being undertaken?
 
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  • #31
russ_watters
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This passed without enough notice:
If this will be in a vacuum later then heat transfer from the environment to the dewar will be greatly reduced. You should take this into account.
At room temperature and pressure, the majority of that 1KW that V50 calculated is via convection. Removing the air removes the convection and the boiling rate will drop significantly unless that power is replaced. @Phys12, this takes you backwards!
 
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  • #32
russ_watters
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The reason why I used the equation of heat capacity is because this is what my actual experiment was going to look like: I would first put my dewar in high vacuum and then fill it with liquid argon. After that, I made a crude approximation to find out the power I may need and assumed that initially the argon wouldn't be boiling at all (since it's not at room temperature as the dewar was closed and evacuated) and I would have to raise the temperature of it in order to make it boil. I didn't go through the process of first calculating the heat it takes to raise the liquid to its boiling point and then using specific heat to figure out how much of the liquid I'm converting to gas in what amount of time given the amount of heat I'm adding...

So, I think that calculation is wrong, but for a different reason. Reason being that I calculated only the heat needed to raise the temperature of the liquid by a few Kelvin, but not the heat required to convert that volume to gas (note that the calculation which yielded 3600 kJ was to raise the temperature of about 100 L of liquid argon to about 10K). I used 10K since the difference in the temperature of liquid argon and its boiling point is about a few Kelvin...
I only considered the part of raising the liquid's temperature to its boiling point...

To be precise, it would have giving me lower heat than I actually need since I first need to heat the liquid to its boiling point (which I calculated) and then convert it to gas (which I missed). I don't understand where I insisted that the wrong equation will give me the right answer...I may have insisted that I do not have a misunderstanding when it comes to knowing the difference between heat and temperature and between latent heat and heat capacity, which I think I'm right about.
I'm at a bit of a loss here, and I promise I'm not trying to be mean, but these statements reflect a near total misunderstanding of the relevant basic chemistry of heating and boiling a liquid. Things that should have been taught in high school.

What you've said you want to do is transfer liquid argon from a (presumably) room temperature and pressure storage tank to a room temperature evacuated dewar. Here's what happens:

First, the environment around the storage tank is room temperature, but the argon is at its atmospheric pressure boiling point of 87K. Principle: The boiling point of a chemical liquid is a function of pressure (higher pressure = higher boiling point). The boiling point has nothing to do with the temperature of the environment around it.

When you transfer the argon to the evacuated dewar, presumably by hose and controlled manually by a fill valve, the argon will have to cool the hose and dewar from room temperature, which heats and boils-off the argon (so your vacuum pump will need to deal with that). Definition: "heat" (verb) means to transfer or add thermal energy to something. Heat (noun: thermal energy in transit) flows from warmer things to colder things, but the objects do not need to be "hot" to transfer heat. You expressed surprise that your resistor was cold when you pulled it out of the liquid nitrogen: of course it was cold - it was just immersed in liquid nitrogen! But that doesn't mean it wasn't doing heating.

Simultaneous to the heating but more specific, the argon pressure drops from atmospheric pressure to near vacuum as it passes through the valve. Per the principle above, the boiling point of the argon *instantly* drops. Due to the mismatch in temperatures, argon now has an excess of thermal energy and a portion of it *instantly* vaporizes ("flashes").

Eventually, this process will reach a steady state where you have:
-Liquid argon flowing from the storage tank at atmospheric pressure.
-Liquid argon flows through the valve, its pressure and temperature drops, and a portion flashes to vapor.
-A liquid/vapor argon mix flows into the dewar where the liquid drops to the bottom and absorbs heat from whatever sources are available (radiation through the vacuum chamber, conduction, an electric resistor in the dewar, whatever) and boils.
-The vacuum pump draws-off very cold, low pressure argon vapor.

As you can see, *nowhere* in this process is the temperature of the liquid argon rising to get to its boiling point. Being *exactly at* its boiling point is its default state.
 
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cjl
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The goal is to recycle argon, while also increasing its purity. Over time, we expect the purity of argon to decrease and we're hoping this to help with that.

The only mechanism I can think of that would cause the purity to decrease though would be that water and oxygen would condense into your liquid argon, due to their higher boiling points and presence in the atmosphere. Unfortunately, though, if this happens, boiling off the argon and trying to purify the vapor that comes off is going to increase the concentration of impurities in the remaining liquid, not decrease it. This will happen because the argon, being the component of the mixture with the lowest boiling point, will preferentially boil off first, leaving the remaining liquid enriched in oxygen and water relative to the initial concentration.

If you really need to keep the argon pure, your best bet is to ensure that oxygen, water, and other impurities don't come in contact with the free surface of the argon in the first place by purging your container with argon after filling. Luckily for you, argon is denser than air, so even if you don't take any special precautions, the gas directly above your liquid should be nearly 100% argon just from normal boiloff as long as you don't have any air currents to disturb it.
 
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