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

 


  1. #101
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    Quote Originally Posted by jimmyw View Post


    No, I think the volume of the gas in the donor tank is 80cf and the volume of the gas in the receiving tank is 0cf



    As awap points out... "The tank is important"

    The "V" in the gas formulas is the volume of the container holding the gas... not how much volume the gas would occupy if released into the atmosphere at sea level.

    If you took the 80cf donor tank to the top of a mountain it does not become a 90cf tank, does it? If you took it into space, would it become an "∞cf" tank?

    The volume of the gas in the donor tank is 11 liters, not 80cf. Same for the receiving tank.

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    I think it's time to ask again if I should just go away? There have been three really good alternative explanations for the heating, 1) residual air compression, 2) heat leakage in at the transfill and valve and 3) the J-T coefficient of non-ideal gas air. They all taught me something as I looked them up and studied them. I find myself almost completely convinced that I understand this and that those explanations, while valid descriptions of the physics, don't address the basic issue. The math works out. The physics explained in every reference I've looked at matches up. The answer seems to be that the transfill operation is "free expansion" - the temperature is basically unchanged of the total gas, but the transfill whip and valve causes energy to separate between the donor and receiving tanks during the expansion - resulting in different temperatures in the two tanks.

    All that's left is some niggling concern that perhaps there's another detail I've missed, but as the posts pile up, I'm less and less convinced there's more to learn, and more and more I find myself trying to convince others. That's not why I came here. Sure, I'd like everyone her to suddenly say "Hey - perhaps he's got a point - let's look at this more closely", but I can live with being thought wrong by others. More and more, I'm moving towards the point of trying to convince others, and that's not what I want to do here. The only reason to try to explain my conclusion is to let others pick holes in it.

    So I end up asking again - Should I just go away? I'm willing to learn and I'm willing to explain what I think is the answer to see why others disagree or agree, but if this thread is becoming offensive, it's time to go.

    Thanks very much for the patience of all and for the time others have spent - whether I agree with their comments or not.

    ---------- Post added February 25th, 2013 at 12:29 PM ----------

    Quote Originally Posted by RJP View Post
    No, I think the volume of the gas in the donor tank is 80cf and the volume of the gas in the receiving tank is 0cf

    As awap points out... "The tank is important"
    The "V" in the gas formulas is the volume of the container holding the gas...
    Good catch. I posted too quickly. My mistake - you are right. The volume of the gas is 11 liters, not 80cf. Sorry for the mistake - I've been typing quickly. I did not use 80cf in any calculations - I always used 11 liters. The volume of the gas is 11 liters and the volume of the tank is 11 liters, and I did say that elsewhere in the post.

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    Quote Originally Posted by jimmyw View Post
    I'm willing to learn...
    Not seeing it.

    Quote Originally Posted by jimmyw View Post

    ...and I'm willing to explain what I think is the answer to see why others disagree or agree,
    That's the one thing you've consistently made abundantly clear.


    Quote Originally Posted by jimmyw View Post
    but if this thread is becoming offensive, it's time to go.
    As mentioned above, it's all in good fun. But at some point you should accept two things:

    1.) You're wrong.
    2.) The receiving tank warms up because the gas in it has been compressed.

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    Quote Originally Posted by jimmyw View Post
    I think of the refrigeration cycle as relating to the transfer of heat energy to/from the environment, but I won't disagree that it would be understood if you understood the refrigeration cycle. Whether that understanding is "easily" obtained is certainly debatable in light of this thread.
    Sorry I should added a few qualifiers.

    Easily understood by those capable and interested in learning and equipped with at least an average IQ.

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    I hesitate to waste more time on this, but... there are some screaming fundamental errors being made. One cannot "do" math without sorting some of this out.


    Regarding these tanks you speak of: 11 litre cylinders. These have a wet volume of 11 litres... about. Let's keep the arithmetic simple and say that we substitute 10 litre tanks instead. Is that OK?

    Empty one will contain about 10 litres of gas at one bar. The charged one will contain 10 litres multiplied by its actual fill pressure... let's say 200 bar. So, there are 10 litres in one and 2,000 litres of gas in the other.

    Now, let's also assume that the gas and the transfill system itself (ALL components) have reached a state of thermal equilibrium... a fancy was of saying they are the same temperature... normally we might cite room temperature, let's say 20 degrees (68 F)... but we are trying to do maths around gas which is basic chemistry so we should work in Standard Temperature and Pressure (STP), and that should be 273 K (0° Celsius) and 100 kPa or one bar. And we can also work out the quantity of gas we have... in MOLES. This is somewhat pedantic but hey, read this thread.


    At STP one mole of gas has a volume of around 22.4 litres. Therefore we have approximately 89.33 MOLES OF GAS... The 2000 litres in one cylinder and the one litre in the other... plus a few ccs in the fill whip. Cool.

    Now we should work out the specific heat capacity AND the dynamic, absolute and kinematic viscosity of the gas we are dealing with (which is what by the way, air?) This is going to give us a better bridge to plugging the correct values into this piece of math we are about to embark upon...


    Hang on. Hang on...

    What was the original question, again?

    why do tanks get hot when you fill them [from] higher-pressure tanks?

    Oh, that's simple... we don't need all this mathematics -- unless we REALLY want to do some heat sink calcs -- we simply need to understand the basic properties of a fluid... in our case a gas.

    But first, let's throw out this Ideal Gas Law crap. Gas has mass and Van Der Waals tells us that when dealing with a fluid's behavior, we have to take into account the nonzero size of a fluid's molecules (or particles) and the attraction between them and interaction with things they come into contact with.

    Now things may start to make sense... if we opened the valve joining the two tanks that are part of our sealed system, we will have interfered, tampered with, altered and changed its state. Essentially, when the dust settles, we will have the same quantity of gas (a little more than 89 moles) but now it will be shared between two containers rather than the bulk being in one. As both sides of the system changed state and reached this altered state (an equal pressure, which for the sake of brevity we can call 100 bar), a couple of other things happened: The pressure in one cylinder dropped and in the other it rose. During this process, there was a transformation of energy... potential into kinetic. During that process, molecules of gas interacted with each other and their surroundings. This got them excited. In the real world we call this excited state: HOT. For a graphical explanation look at a picture of Nicole Kidman and imagine her moving around excitedly in close proximity.

    This is a simplified explanation, but surely it serves your purpose. ALL the rest of this twattle is simply that... rubbish punctuated with inaccurate assumption and shady science. No need to go away... no need to shut up... but if you want more, you WILL NOT find it here. Actually, you have received some very valuable responses, but you seem intent on nit-picking. You need to read and understand the gas laws, thermal dynamics and use proper scientific units... as a starting point. It gets more complicated after that.
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    Quote Originally Posted by Doppler View Post
    . In the real world we call this excited state: HOT. For a graphical explanation look at a picture of Nicole Kidman and imagine her moving around excitedly in close proximity.
    Just when I thought my interest in this discussion had started to wane...

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    If I may...I'm called upon many times (per year) to explain this phenomenon to freshman science students. I've found a conceptual understanding before attacking the equations and formulas useful. It's rudimentary, but it seems to solve the problems people have with understanding what is happening.
    I start with the kinetic theory, which states that all matter is composed of tiny particles (molecules or atoms) that are in CONSTANT motion. The higher the KE of the particle, the faster it moves and vice versa. So, that is what seperates the phases...gases have higer KE of particles than say, liquids or solids. Now, on to point two...
    When a gas is compressed into a fixed volume (such as a scuba tank), the particles ability to move is restricted, therefore, the KE of each particle is REDUCED. Since energy cannot be destroyed, only converted, the KE of the particle must take another form. In this case, the KE is converted to heat which is radiated and conducted outward into the metal of the tank. The tank being filled gets hot.
    Conversely, for the tank that is being used as the filler, as the gas is moved from higher pressure to lower pressure, the particles expand, thus gaining KE. That energy must come from somewhere, and in this case it comes from anything it touches in the process of expanding...the tank, the hose, any other gases it with which it comes into contact, etc...therefore that tank COOLS.
    Yes, I know this is simplistic, but it seems to build a conceptual model that my students can absorb before we start crunching numbers. Hope it helps.

    ---------- Post added February 25th, 2013 at 12:47 PM ----------

    Dang, Doppler...you beat me to it while I was composing my response! Good job, though!
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    Quote Originally Posted by Doppler View Post
    I hesitate to waste more time on this, but... there are some screaming fundamental errors being made. One cannot "do" math without sorting some of this out.
    I'm game - let's do it

    Regarding these tanks you speak of: 11 litre cylinders. These have a wet volume of 11 litres... about. Let's keep the arithmetic simple and say that we substitute 10 litre tanks instead. Is that OK?
    Great - yes 10 litre tanks.

    Empty one will contain about 10 litres of gas at one bar. The charged one will contain 10 litres multiplied by its actual fill pressure... let's say 200 bar. So, there are 10 litres in one and 2,000 litres of gas in the other.
    If you insist on keeping air in the second tank, OK. If you can do it with a vacuum in there, the math is easier, but I'll do it your way.

    Now, let's also assume that the gas and the transfill system itself (ALL components) have reached a state of thermal equilibrium... a fancy was of saying they are the same temperature... normally we might cite room temperature, let's say 20 degrees (68 F)... but we are trying to do maths around gas which is basic chemistry so we should work in Standard Temperature and Pressure (STP), and that should be 273 K (0° Celsius) and 100 kPa or one bar. And we can also work out the quantity of gas we have... in MOLES. This is somewhat pedantic but hey, read this thread.
    OK

    At STP one mole of gas has a volume of around 22.4 litres. Therefore we have approximately 89.33 MOLES OF GAS... The 2000 litres in one cylinder and the one litre in the other... plus a few ccs in the fill whip. Cool.
    Remember, the "2000 litres in one cylinder" are at a totally different pressure from the receiving tank, and you said "one litre in the other" when I think you meant 10 litres in the other.

    I'm going to work out the numbers this way -
    We have 10 liters of air at 200 bar and temp 273 K (0° Celsius) .... I got to this point and then realized you didn't actually do the math. I have done the math. Repeatedly. Is it worth going through it? To do that, I have to state whether we are adding energy to the gas or not. That's why I asked these questions:

    1) Did it take energy to compress the gas into the donor tank?
    2) Could we get some of that energy back by putting a pneumatic motor/generator in the whip line between the tanks and driving the generator by the pressure differential during the fill?
    3) Would the recipient tank and donor tank equalize pressure regardless of whether we put a pneumatic motor/generator in the whip line?
    4) What happens to the energy we could have extracted from the gas with the pneumatic motor/generator if we don't put the pneumatic motor/generator in the whip line?

    Are they that hard?

    Let me give the answers I think are pretty clear. If my answers are wrong, tell me which ones are wrong. At least give me some hints on where you think this is wrong:

    1) Did it take energy to compress the gas into the donor tank?
    A: Yes. We have to do work on the gas to push it into the donor tank. The compressor does that work. As a side note, this is called a reversible adiabatc process of constant entropy. The internal energy of the gas is raised as it goes from 80cf at STP to a higher temperature pressure and smaller volume. Because this process is "reversible" (another way of saying the entropy is unchanged), we could extract the stored energy by letting the gas expand and do mechanical work on a turbine/generator. The gas would cool if we did that and return to STP. The cooling would exactly counteract the heating from the original compression. The extracted energy could be used to reverse the process again and push the gas back into the donor tank, which would heat it up. In theory, we can go back and forth - compress hot - expand cool. Below I will say we could let the expanding gas out into a tall vertical cylinder with a weight on a piston. We could let it push the cylinder of mass m upwards to a height h and store mgh potential energy. We could then push the cylinder and weight downward to compress the gas and make it hotter. We can go back and forth (neglecting friction).

    2) Could we get some of that energy back by putting a pneumatic motor/generator in the whip line between the tanks and driving the generator by the pressure differential during the fill?
    A: Yes. As described above. We don't have to use a pneumatic motor and generator. We could let the gas expand into a cylinder and push a weight on a piston upwards and use mgh where m is the mass of the rising piston and h is the height the gas pushes that weight up. It doesn't matter how we extract the energy or how we store it. What is important, however, is that if we *do* extract the energy, all the gas in the donor tank, including the gas we let out of the donor tank into the cylinder not only expands, it also cools. Again this is a reversible constant entropy process. I'm sorry I'm using technical terms, but they accurately describe what's happening.

    3) Would the recipient tank and donor tank equalize pressure regardless of whether we put a pneumatic motor/generator in the whip line?
    A: Yes. If we let the gas push a mass upwards, it's a reversible process, and we extract energy, but if we were to just let the gas into an empty cylinder, with the weight already at the top of the cylinder, it would be an irreversible, process. If I can be excused for using another technical term, this type of expansion is called constant enthalpy. The constant enthalpy process is not reversible, but has constant internal energy of the gas. Either way, the final pressure would be the same in both tanks, but in one case, the temperature of the gas would decrease and in the other (called "free expansion") the temperature would be unchanged in the "Joule expansion = free expansion" process - provided we let the gas freely mix. If we don't let it freely mix, some gas is hotter and some gas is colder.

    4) What happens to the energy we could have extracted from the gas with the pneumatic motor/generator if we don't put the pneumatic motor/generator in the whip line?

    A: The energy that could have been extracted is released as heat into the gas in the receiving tank. Instead of the expansion cooling the gas in the receiving tank, as happened in the reversible process, it is released as heat, and there's just enough energy to counteract the cooling of all the gas from the expansion that would have occurred if we'd let the expansion raise the weight in a reversible process.

    I can do the math on all of this. I can start with 80cf of air at STP and calculate how hot it gets when we compress it reversibly. If preferred, I can let the excess heat escape and figure out the pressure (PV is constant since T is constant) I can then assume no energy (no work no heat) is added to the gas and determine the final temp during free expansion. It will be room temp if all the gas is allowed to mix and equalize temp. I can also calculate how hot the gas in the receiving tank gets if we assume that the gas in the donor tank expanded reversibly, and all the released energy is transferred to the gas in the receiving tank. I can even calculate what happens if there is residual gas in the receiving tank. And if necessary, I can even add in the final details of assuming that the gas is not ideal so that the J-T coefficient is non-zero.

    It's long and boring, and I suspect that it won't convince anyone, but it convinced me.

    If anyone wants to understand the physics, they have to figure out the difference between extracting energy from the compressed high pressure gas in the donor cylinder versus not extracting/storing that energy and letting it appear as heat in the receiving tank gas. All of this works for the ideal gas assumptions, and for real world gases.

    Here's another example - we let the gas from the expanding donor tank push a bullet through the whip hose. The bullet has mass and picks up energy, then slams into the receiving tank and it's kinetic energy converts to heat. It's the same as letting the expanding gas of the donor tank push and accelerate the gas leaving the donor tank, but it might be easier to see it that way.

    Last edited by jimmyw; February 25th, 2013 at 04:00 PM.

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    Jimmy,

    Nobody cares. All the questions of merit have been answered.

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    Quote Originally Posted by Guba View Post
    If I may...I'm called upon many times (per year) to explain this phenomenon to freshman science students. I've found a conceptual understanding before attacking the equations and formulas useful. It's rudimentary, but it seems to solve the problems people have with understanding what is happening.
    I start with the kinetic theory, which states that all matter is composed of tiny particles (molecules or atoms) that are in CONSTANT motion. The higher the KE of the particle, the faster it moves and vice versa. So, that is what seperates the phases...gases have higer KE of particles than say, liquids or solids. Now, on to point two...
    When a gas is compressed into a fixed volume (such as a scuba tank), the particles ability to move is restricted, therefore, the KE of each particle is REDUCED. Since energy cannot be destroyed, only converted, the KE of the particle must take another form. In this case, the KE is converted to heat which is radiated and conducted outward into the metal of the tank. The tank being filled gets hot.
    I agree with this. Compressing a gas heats it up as you add energy by doing work on the gas. The internal energy of the gas is increased.

    Conversely, for the tank that is being used as the filler, as the gas is moved from higher pressure to lower pressure, the particles expand, thus gaining KE. That energy must come from somewhere, and in this case it comes from anything it touches in the process of expanding...the tank, the hose, any other gases it with which it comes into contact, etc...therefore that tank COOLS.
    Consider "free expansion" also called "Joule expansion" as described here. The cooling always occurs during a reversible expansion, but for constant internal energy, constant enthalpy expansion, there is no cooling for an ideal gas and only small differences for non-ideal gases.

    I welcome your thoughts. The transfill operation is very close to the described free expansion/Joule expansion case, except that the gas is not mixing freely. The gas in the donor tank is closer to a constant entropy expansion, so it cools, and the released energy appears as heat in the receiving tank. At least that's how I interpret it.

    Guba, as a science teacher, I ask you to look over my answers to the four questions above, or answer them yourself. If I'm off target, give me some help on getting back on track. I can't find any errors.

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