Why do tanks get hot when you fill them from higher pressure tanks?

[RJP]

Can you repeat the question? hehe....

"Is this a five-minute argument, or the full half-hour?"

[video=youtube;hnTmBjk-M0c]https://www.youtube.com/watch?v=hnTmBjk-M0c[/video]



---------- Post added February 23rd, 2013 at 03:12 PM ----------

Strangely, I enjoyed reading this thread. I am not sophisticated about th physics of tank filling, but I have filled thousands of tanks from a bank. Here is some observational data for the scients to incoprate in their continued explanation:
1. The speed at which you fill a tank impacts the temperature of the receiving tank ( and the air in it). The faster you fill, the hotter it gets.
2. As the "hot" tank cools, the pressure reduces due to cooling of the contents, resulting in what was a 3000 PSI fill ending up being a 2400 psi fill ( more or less). The full cooling of the filled tank will take an hour or more on a hot summer day, even if in the shade.
3. If you fill a tank at about 500 psi per minute or less, the heating is minimal, but it takes a long time to fill 50 tanks after the day's dive boat trip is done.
4. If you place the receiving tanks in cold water as you fill them, you can fill faster, as the heat dissipates quicker in the cold water ( just like the heat from a submerged diver's body).

So, if anyone cares to assay this question, please explain the relationship between rte of fill and temperature of air in the recipient tank.
DivemasterDennis

Heat is caused by the act of "compression" not by the steady state of "being compressed." This is why tanks cool and don't stay hot. Think of compression in terms of the aforementioned "crashing." Heat is the release (transfer, really) of energy as the result of those crashes. The faster that tanks are filled the faster you cause gas molecules to crash into each other. Not just in terms of the speed of each individual crash, but the cumulative rate at which all of these individual crashes pile up.

Think about a car accident. One car bumping another in a parking lot at 5mph will not release much energy. Even a row of cars in a chain-reaction bumper-to-bumper collision inching towards a toll-booth won't do much damage. But two cars crashing into each other at 50mph will release a lot more energy.

Now imagine a fifty-car pile up at highway speed.

 

[jimmyw]

I have a dilemma. I think I now understand the physics of what is going on. I seem to be in the minority, but I suspect that with careful explanation, I can explain it well enough to convince anyone who is interested that the physics is what it is. I'm willing to try. At the same time, I don't claim to be an expert. If I'm wrong, I'd really like to know. I'd like to be able to understand this, so if I am wrong and I'm missing something, I want to know it. The only way that's going to happen is for me to consider other viewpoints and really think through what others have to say on the subject.

However, at the same time, I don't want to just engage in a pointless set of arguments. I'm not here to cause trouble, just to learn and to try to arrive at the truth. I'm new here and don't feel I have the right to undermine the reputation that others have here. So I'd like some feedback. I've made most of the points that I think point to the right answer, but I haven't posted any mathematics or the gas law analysis that underlie my conclusions. That would be the logical next step.

Should I just go away, or are the other divers here prepared to bear with me while I work through all the objections and convince myself (and hopefully at least a few others) that I have the physics right? I promise to be polite. To the best of my ability I'll base my comments purely on the physics and math of the gas laws. I may have to delve in a bit more deeply into entropy, enthalpy, internal energy of gas and other physics of gases if I'm to go beyond what I've already posted. I've done that for myself and the more I've done it, the more it's become clear to me what's going on.

I understand that no one has to read this thread if they don't want to, but I don't want it to degenerate into something that's really an objectionable thread to read. If you want me to disappear, now is the time to say so. If you'd rather listen to what I have to say, and hopefully respond to me to either explain to me why I'm wrong or be convinced that I really do understand what's happening, then say that.

Thanks for listening to me so far, and as always, be safe.

 

[RJP]

I have a dilemma. I think I now understand the physics of what is going on. I seem to be in the minority, but I suspect that with careful explanation, I can explain it well enough to convince anyone who is interested that the physics is what it is. I'm willing to try. At the same time, I don't claim to be an expert. If I'm wrong, I'd really like to know.

Give it a go, this is all in good fun. However, and don't take this the wrong way, do you intend to try to convince yourself of what's actually happening? Or is it still your desire to convince those with an understanding of both the theory and the practical nature of the situation that your incorrect understanding should actually supplant proven science and empiric observation?

:d

I'm happy to play along, either way. I had a roommate in graduate school whose thesis sought to prove why bats can't fly. It was a ton of fun to watch.

So I'd like some feedback. I've made most of the points that I think point to the right answer, but I haven't posted any mathematics or the gas law analysis that underlie my conclusions. That would be the logical next step.

I may have to delve in a bit more deeply into entropy, enthalpy, internal energy of gas and other physics of gases if I'm to go beyond what I've already posted. I've done that for myself and the more I've done it, the more it's become clear to me what's going on.

But seriously, I think "the logical next step" for you should be to walk through where in the system gas expands or is compressed or might change temperature during transfill and equalization of two tanks. I believe that is where your problem lies. Just list them out. Once you've got a handle on that, you merely apply the laws of physics. The fact that you've stated several times that "the gas expands into the receiving tank" tells me you've got a fundamental misunderstanding of the practical aspects of the situation, not gas laws and equations.

 

[halocline]

Jimmy, is your question really as simple as "why does a tank heat up when being filled from another tank?" I was only sort-of kidding when I asked if you could repeat the question.

RJP, I gotta hand it to you, you're patient and persistent.

 

[jimmyw]

Jimmy, is your question really as simple as "why does a tank heat up when being filled from another tank?" I was only sort-of kidding when I asked if you could repeat the question.

Yes, it's as simple as that. Except that I see a big difference between why it gets hot when it's filled from a compressor and why it gets hot when it's filled from a high pressure tank. When running a compressor, we are taking power from the wall to run the compressor. When filling from a donor tank, we're not doing that. We're just letting the gas expand out of the donor into the receiving tank. One seems to be creating heat from electrical energy from the wall, while the other seems to be making heat from nowhere. At least that's what it looked like at first. They just seem to be totally different to me.

To answer RJP's question. I want to understand what's going on and I believe that there is only one real answer that we should all be able to agree on. My goal is not to convince anyone else of the answer, but only to understand what the answer is, and, if others disagree, to figure out if they are right or wrong. I don't mind if I've come to the wrong answer, but if I have, I want to know why I'm wrong. If I have the right answer, I don't mind if others disagree, as long as I've convinced myself that they are wrong. When I started this, I was right in the middle. I felt something was odd, but couldn't put my finger on it. Now, I think I do have the answer, but since I seem to be mostly alone, there's got to be a good chance I'm wrong. If I am, however, I can't find the error in the physics.

So, let me see if I can address this part of RJP's post:

The fact that you've stated several times that "the gas expands into the receiving tank" tells me you've got a fundamental misunderstanding of the practical aspects of the situation, not gas laws and equations.

I agree this is where we disagree.

Let's try to reach some sort of agreement on this first. It seems clear to me that the gas that goes into the receiving tank comes from the gas in the donor tank and that gas started at higher pressure.

Can we agree that the description of "Joule expansion" in the Wikipedia article is basically the same as what happens when we fill a receiving tank from a donor tank? To make it as similar as possible, let the donor tank and receiving tank be identical in size. Let the receiving tank be empty when we start and let the donor tank and receiving tank pressures equalize. The only difference I see is that the Joule expansion case allows the gas in both tanks to freely mix so that all the gas ends up at the same temperature.

I quote from that article below.

Joule expansion is an irreversible process in thermodynamics in which a volume of gas is kept in one side of a thermally isolated container (via a small partition), with the other side of the container being evacuated. The partition between the two parts of the container is then opened, and the gas fills the whole container. This process is a useful exercise in classical thermodynamics, as it is easy to work out the resulting increase in entropy, the so-called entropy production.
...
Joule expansion is also called free expansion.
...
Description

We consider n moles of an ideal gas at pressure P_i and temperature T_i, confined to the left-hand side (as drawn) of a thermally-isolated container, that occupies a volume V_i = V₀. The right-hand side of the container, also with volume V₀, is evacuated. The tap (solid line) between the two halves of the container is then suddenly opened and the gas fills the entire container of volume V_f = 2V₀. We propose that both the previous and new temperature and pressure (T_f, P_f) follow the Ideal Gas Law, so that initially we have P_iV_i = nRT_i and then, when the tap is opened, we have P_fV_f = nRT_f, where R is the molar ideal gas constant.
As the system is thermally isolated, it cannot exchange heat with its surroundings. Also, since the system's volume is kept constant, the system does no work on its surroundings.^[3] As a result, the change in internal energy ΔU = 0, and because U is a function of temperature only for the ideal gas, we know that T_i = T_f. This implies that P_iV₀ = P_f(2V₀), and thus the pressure halves; i.e. P_f = ½P_i.

If we can't agree on this, then perhaps we should go back to the case we do agree on - directly filling from a compressor and analyze that case. That case involves adding energy to gas. The energy is in two parts - work, which is the force applied to push the gas into the tank times the distance we have to push it, and heat, which is the adiabatic compression heating we are all familiar with.

If we go back to the start - when the donor tank was filled - we can then track that energy in the gas as it leaves the donor tank and enters the receiving tank.

---------- Post added February 24th, 2013 at 09:47 AM ----------

Edit: If anyone agrees with me that the donor/receiving tank and the Wikipedia Joule expansion/free expansion cases are the same (except for the gas mixing), I'd appreciate it if they would say so. OTOH, if they disagree, I'd appreciate it if they would say that and explain why they disagree. It does seem to be close to the heart of the issue.

 

[RJP]

Edit: If anyone agrees with me that the donor/receiving tank and the Wikipedia Joule expansion/free expansion cases are the same (except for the gas mixing), I'd appreciate it if they would say so. OTOH, if they disagree, I'd appreciate it if they would say that and explain why they disagree.

It does seem to be close to the heart of the issue.

The heart of the issue continues to be that you really have no idea what's going on here. Seriously. No one can explain "why they disagree" with you, other than to point out that you are wrong. As a logic professor in college used to say "Error has no defense."

No way I'm reading everything you just posted above closely, because it starts out wrong from the get go...

FIRST: The air coming out of the compressor is heated because is has been compressed. Has nothing to do with taking heat from the electricity coming out of the wall. I could run the compressor in a walk-in freezer hooked up to an exercise bike and the air that is compressed would still be warmer than the ambient temperature at which it started. Go grab a bicycle pump, hook it up to a tire, give it one compression. Feel the wall of the pump. It's warm. No electricity, no engine, no nothing. Just compression. Yes, energy from your arms was expended. But the heat is created from the compression of the gas, not the warmth of your muscles.

SECOND: The air does NOT NOT NOT NOT NOT "expand into the receiving" tank. The air is COMPRESSED into the receiving tank. What causes this compression? The pressure of the gas remaining in the donor tank pushes the gas into receiving tank until the two tanks equalize. Again, if this was not the case, you would not be able to transfill the tank, because you would not be able to get more than 1atm/14psi into the tank without compressing the gas. I just ran this concept past my 8yr old son. His response? "Duh..."

Because the air is compressed into the receiving tank it warms up. Just that same as if it came from a compressor, a bike pump, or any other manner you can find to get more than 1atm of air into the tank. In your words above, the pressure gradient of the gas in the donor tank "adds energy" to the gas in the receiving tank. In fact, the 3,000psi of energy stored in the donor tank came from a compressor at some point. That "compressor energy" is stored their as 3,000psi of pressure. When you open the valve that energy - that was initially transferred to the donor tank from a compressor - is then transferred to the receiving tank.

You're so worried about entropy and enthalpy and formulas that you're completely missing the obvious, fundamental flaw in your thinking: You believe that the gas going into the receiving tank is expanding. It is not. It is being compressed. Your second issue, that somehow an electric/gas compressor somehow heats gas by a means other than compression in a way that is not possible by other means is also flawed.

 

[CT-Rich]

A simple way to look at this is to think of each molecule moving around with x amount of kinetic energy. When a molecule hits the side of the cylinder, some energy is transfered to the cylinder. The energy of the individual molecules would not have to change, however the more molecules the more frequent the collisions. As you increase the pressure on chamber, you are increasing the total energy contained in the gas within a fixed volume, heating the tank.

The reverse happens when you release gas from a cylinder. The release of molecules out of the nozzle is releasing the kinetic energy of the gas into the surroundings and the temperature of the cylinder goes down. This the basic principle for most A/C and refridgerators.

When you transfer gas from a bank of bottle to a empty scuba cylinder, the total change in energy is zero. the increase in temperature (total kinetic energy of the molecules) cancels out the drop in temperature of the gas bank. The bank is larger than the single cylinder so it is less noticable.

 

[RJP]

A simple way to look at this...

Yeah. Good luck with that.

 

[redacted]

Both are correct. But the Joule Expansion is only theoretically possible. It does not reflect the reality of your question. That is, when filling one scuba tank from another, you do not meet the condition of thermal isolation necessary to produce the described results. Sure, a compressor adds a thermal source but even with bank tanks, the event is not thermally isolated and adiabatic heating and cooling occur and the gas molecules take up and give of energy as the pressures change during the fill process.

 

[jimmyw]

Strangely, I enjoyed reading this thread. I am not sophisticated about th physics of tank filling, but I have filled thousands of tanks from a bank. Here is some observational data for the scients to incoprate in their continued explanation:

Thanks for the observations, Dennis. I'm going to comment on these points so that I can explain to myself why I think these things happen. I'm doing this in order to help me spot any errors in my conclusions, or possibly to let others spot those errors. I think I can explain all the observations consistently, and I think my comments wil also provide some answers to others who have urged me to track each step or stage of the filling process and apply the gas laws at each individual step or stage of the process.

1. The speed at which you fill a tank impacts the temperature of the receiving tank ( and the air in it). The faster you fill, the hotter it gets.

I think this happens because the faster you do it, the less time there is for the hot tank to cool. If we clamped both tanks together and wrapped the entire thing in insulation, I've calculated that the whole thing (both tanks and all the connecting valves, etc.) would all end up at room temp, no matter how quickly the transfer was made.

2. As the "hot" tank cools, the pressure reduces due to cooling of the contents, resulting in what was a 3000 PSI fill ending up being a 2400 psi fill ( more or less). The full cooling of the filled tank will take an hour or more on a hot summer day, even if in the shade.

I think this is just P/T=constant. As T drops, P also drops. It comes from the ideal gas law PV=nRT where volume is constant.

3. If you fill a tank at about 500 psi per minute or less, the heating is minimal, but it takes a long time to fill 50 tanks after the day's dive boat trip is done.

Same answer as for 1 above.

4. If you place the receiving tanks in cold water as you fill them, you can fill faster, as the heat dissipates quicker in the cold water ( just like the heat from a submerged diver's body).

This makes sense for the reason you give.

So, if anyone cares to assay this question, please explain the relationship between rte of fill and temperature of air in the recipient tank.

If you look at the gas laws - ideal gas law, Charles's law, Henry's law, etc. you'll see that none of them have any "time rate of change" factor. We can "calculate" the final temperature, pressure and volume without knowing the rate at which the changes were made.

However, all the fill operations we do aren't perfect the way those laws are perfect. They assume no energy gets lost, while in the real world, hot things cool off, so the longer we take to heat them up, the lower the final temperature.

---------- Post added February 24th, 2013 at 10:40 AM ----------

When you transfer gas from a bank of bottle to a empty scuba cylinder, the total change in energy is zero. the increase in temperature (total kinetic energy of the molecules) cancels out the drop in temperature of the gas bank. The bank is larger than the single cylinder so it is less noticable.

I agree with this.

---------- Post added February 24th, 2013 at 11:06 AM ----------

Both are correct. But the Joule Expansion is only theoretically possible. It does not reflect the reality of your question. That is, when filling one scuba tank from another, you do not meet the condition of thermal isolation necessary to produce the described results.

We agree that the real world isn't perfect, but no temperature change - Joule expansion was shown experimentally. If the receiving tank is thermally connected to the donor tank and the whole thing is wrapped in insulation we should have the same process.

I read several descriptions of the Joule expansion experiments. In one, they put the receiving tank inside the donor tank, then opened the connecting valve. The gas arriving in the inner receiving tank got hot. The gas remaining in the outer donor tank cooled off. The whole thing was wrapped in insulation to prevent any heat getting to the cold outer tank. IOW, the energy added or removed from the system was zero. When the gas of the inside tank stopped cooling and the gas of the outside tank stopped being heated by the inside tank, the temperature of both tanks was the same and that temp was unchanged from the original temperature.

That tells me that the fundamentals of filling a tank from another tank are the same as in Joule expansion/free expansion.

Sure, a compressor adds a thermal source

I was writing quickly and didn't mean to imply that a compressor adds any heat energy. (In the real world it does, but that's not where the heat comes from) Energy is work plus heat energy. The compressor is doing "work" - applying a force to the gas and moving it a distance to force it into the tank.

We can assume that all the energy added is in the form of work - not heat. Then we just need to solve PV=nRT to get the new temperature and pressure knowing the change in volume - the starting volume of air (80 cubic feet) and the final volume (the tank's true volume). To do that, we need to solve for two unknowns, P and T, but the ideal gas law is only one equation. The other equation I referred to in my first post - (P times V to the power gamma) is constant, where gamma is about 1.4 for air. There's no heat being added, just mechanical work energy, when the donor tank is being filled. I put in those two equations in my very first post to try to draw out the expertise of someone who had looked at the "fill from a higher pressure tank" process from a similar science/physics/math perspective.

but even with bank tanks, the event is not thermally isolated

I agree it's not normally thermally isolated, but the whole effect doesn't go away as we wrap insulation around everything. If anything, the effect becomes more pronounced.

and adiabatic heating and cooling occur and the gas molecules take up and give of energy as the pressures change during the fill process.

We agree that different parcels of gas go through different PVT changes during the process, but as long as we know that no heat is added and no work is done on the system, then we know that the system energy is unchanged. As long as energy is unchanged, all we need are the final pressures and volumes to determine the final temperatures.