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Please Help Me With My Invention i.e. Is This Calculator Really Right?

  1. Jan 27, 2008 #1
    Hello I'm new here. The reason why I have joined is because I'm working on an invention which I believe will be extremely good for the environment. The question is by how much?
    My problem is I never paid much attention at school; lessons I’ll be remembering forever include the one, where in biology we had to dissect sheep’s eye balls and mine was the first to go in the rat cage, or when my chemistry teacher got pissed after we put pure ethanol in his coffee (guess it really is tasteless!).
    And now of course it’s all come back to haunt me, because many formulas, (and most of maths) seems looks more like ancient Egyptian to me, and I have to work really hard to “decode” it.

    The Question…
    If a open cubic meter of air at 1 atmosphere is heated from 300 Kelvin to 400 Kelvin (i.e. from 28 to 128 degrees) its volume will expand by 33.33%
    So logic dictates that if an enclosed meter is heated the same amount then its pressure should increase by 33.33% (i.e. to 1.33 atmospheres)
    http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/ThermProps.htm
    (under “Gas Pressure Increase With Temperature”)

    BUT if I THEN let the pressure DECREASE to 1 atmosphere (i.e. let the gas volume increase by 33.33%) what happens to the temperature?

    At first you would have thought the temperature would decrease by 33% (i.e. from 400 Kelvin to 277 Kelvin) because pressure and volume are supposed to be proportional.
    Here is just one link amongst many that say that
    http://solospirit.wustl.edu/solospirit2/education/Science Side/pvt.html

    But then I came across this calculator…
    http://www.1728.com/combined.htm

    And entered (under the temperature section) the following…
    1. Pressure 1 Equals: 1038.2
    2. Volume 1 Equals: 1
    3. Temperature 1 Equals: 400
    4. Pressure 2 Equals: 760
    5. Volume 2 Equals: 1.366

    Calculators answer: Temperature 2 = 399.98
    (P.S The pressure is in Torr and there are 760 to the atmosphere)

    Obviously online calculators can be wrong, but if a 100 degree temperature increase can give an unenclosed gas volume an increase of 33%; then if enclosed this gas should still be 100 degrees hot after a 33% pressure decrease.
    Then again pressure and temperature are supposed to be proportional.

    So which is right? The calculator, the proportional saying, or even both?
    This problem really bugs me because it will screw up the design in my patient application should I choose the wrong path. So any help would be very gladely received.
     
  2. jcsd
  3. Jan 28, 2008 #2
    It would be helpful if you can describe the processes by which you are doing this.

    This sounds something like an energy source. Is that correct?
     
  4. Jan 28, 2008 #3

    Mech_Engineer

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    Well, tyring to base your patent application off of an online calculator is a bad idea.

    Edit- also, why did you put 1.366 in for the volume of state (2), not 1.333?

    It's true that if you heat a gas in a fixed-volume container the temperature and pressure will rise proportionally (for a perfect gas). It's also true that when you then let the perfect gas expand it's pressure and temperature will decrease proportionally. Assuming of course no heat is put in or taken out of the system during the process (perfect insulation).

    You can't get free energy by heating a gas in a fixed volume and then expanding it. You start at state (1) and end at state (2), no extra energy available. Sorry.
     
    Last edited: Jan 29, 2008
  5. Jan 28, 2008 #4
    Mech Engineer: Don’t worry about the conservation of energy.

    A cubic meter of air weighs about 1.2 kg
    And at standard conditions it has a specific heat of almost exactly 1 KJ for 1KG
    So to heat a cubic meter up by 100 degrees costs 120KJ
    The increase in pressure caused is 33% and the increase in pressure is only worth about 33KJ (and that’s assuming you can recover it at 100% efficiency which of course you can’t).
    The other wasted 67% must be lost to the fact it costs energy to heat air molecules, in the same way it costs energy to heat bricks (even though no useful work is done in the case of bricks).

    Even so (in its industrial context) my idea would save many millions of tonnes of carbon, and it’s clearly MASSIVELY cost effective regardless of what the answer to my question actually is, or if I skip this extra component on the grounds of being slightly uneducated.
    However the problem with this is the first person with just a bit more knowledge than me to study my patient will think; ah!! why not a few cubic meters of air?
    And then my idea will become obsolete, even if the new one is about 90% the same.
    Besides it’s such a shame to ask people to back something that you know could be better-simpler.

    I arrived at your answer too; but the problem it raises is: What’s the difference between the air that Comes Out of a cubic meter of air which is…
    1. 1M3 Heated to 100 degrees in open space, and which therefore expands by 33%
    2. 1M3 heated by 100 degrees in an enclosed space, which therefore causes a pressure increase of 33% but which is then is deflated by 33%.

    Both cubic meters seem to be requiring the same amount of energy to heat them to the 100 degree mark, and both seem to be finishing with the same increase in volume.
    If that’s so, then one would expect the temperature of the air that comes out of them to be the same since energy cannot be destroyed.

    Of course the last cubic meter probably does work even if it’s not useful.

    But wouldn't it be mighty strange to have two 1.3 M3’s of the same mass of gas, under the same atmospheric pressure, but one at a rather different temperature to the other?
    How could that be? After all it would be like the atomic mass had just fallen a few floors through the periodic table.

    In which case maybe the gas calculator is right?
    But if it is right then we run into the original “problem” of there being a decrease in pressure without their being a decrease in temperature.

    I speculate: This might be possible on the grounds of: Pressure and temperature being interchangeable, volume and pressure being interchangeable, but NOT temperature and volume (well unless there’s an explosion).

    Maybe (if the calculator is right) this is why it’s right. Perhaps there is such a thing as a pressure decrease without a temperature decrease if…
    The decrease in pressure, is responsible for a increase in volume, that is under about the same pressure as the new pressure.

    However I don’t know that and wouldn’t be surprised if it was right or wrong.
    So thanks, but I'm still uncertain. So would gratefully like to know more.


    TVP45 Obviously I’m trying not to reveal too much; but what problems is my description causing your thoughts?
    Just in case: Pretend everything is done at 100% efficiency as it’s easy for anyone to estimate, and deduct the real world inefficacies afterwards.
    But to answer your question as best I can: The problem at hand is for one component of many. This components job is to use the increase in pressure caused by heating cold air up to do (a bit) of work. Electricity has problems, so it's been my good fortune to find a niche.

    Anyway thanks for all your help so far
     
  6. Jan 29, 2008 #5
    I was just trying to figure out what your heat cycle looked like, but I think Mech Engineer figured it out for me. Thanks.
     
  7. Jan 29, 2008 #6

    Mech_Engineer

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    I have to say, it doesn't bode well for your invention if I have to ignore conservation of energy to understand it... You need to look at the big picture, where in the start and end state for each process has the same amount of internal energy (indeed, they have the same exact specific volume and pressure as well). The isothermal expansion process you are seeing is because the ideal gas law is a simplified model of gases, in the expansion pressure is decreasing and specific volume is increasing. Let me reiterate, you cannot expect to base a patent application on an online calculator!!! You need to understand the physics behind what you're proposing, and if you do the footwork you'll discover you haven't really "dicovered" anything at all...

    Your major problem is that you are trying to invent a turbine power cycle (or perhaps a piston-based power cycle) using ideal gas law as your guide. Without at very least a basic background in Thermodynamics, this can be an impossible task.

    I suggest you read about the Brayton Cycle, the approximation for a power cycle in a turbine (working fluid is commonly air). The ideal Brayton cycle is broken into 4 steps:

    (1 - 2) Gas is compressed (Isentropic Process)
    (2 - 3) Heat is added to the gas (Isobaric Process)
    (3 - 4) The heated gas is run through a turbine, creating work (Isentropic Process)
    (4 - 1) A small amount of waste heat is exchanged to the environment (Isobaric Process)

    You can see that the Brayton cycle uses terms such as Entropy to describe the gas's properties. As it turns out, a complex power cycle requires more that just basic volume, pressure, and temperature to describe it.
     

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    Last edited: Jan 29, 2008
  8. Jan 29, 2008 #7
    i don't see the problem. P1V1/T1=P2V2/T2. 1.333x1/400 = 1 x 1.33/T2,

    therefore T2=400 as the calculator indicates
     
  9. Jan 29, 2008 #8

    Mech_Engineer

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    The problem is that Out Of Space is having trouble imagining how a gas can go through an expansion while keeping the same temperature unless you add energy to the system. But temperature, pressure, and specific volume are ALL linked, so if one variable is changed and another held constant, the last one remaining will change in the inverse.

    In reality there isn't any energy that's being added to the system through the expansion, but there is an entropy change. When the specific volume changes the pressure changes in response, but the temperature ends up the same.
     
    Last edited: Jan 29, 2008
  10. Jan 29, 2008 #9
    Thank you very much!!!
    Thanks to regor and Mech Engineer I can be confident the calculator is right.

    Mech Engineer I know I can’t base the invention on the online calculator because for one it doesn’t include entropy. However when you look at the answers this one gives you, and when you take entropy into account the answers do make sense.
    But this meant the calculator would be right about there being a pressure change without a temperature change. I was just having trouble confirming this because I can’t read-understand normal calculations very well, instead I tend rely on pictures and volume changes in my mind, and use only a bit of maths (such as pressure = force divide area). It actually works for me very well until I’m left wondering how to confirm something, I'm unsure about.

    Thanks for the Bryton Cycle link. I hadn’t heard of that and so couldn’t Google it; but yes what this component is doing has surely been done before, it’s, its context in the other stuff I daren’t mention that’s new-or unique.
    Basically there is a industry which is currently wasting vast quantises, of fairly high level heat because it would never be cost effective to use water containing pipes to make steam. This is partly because they would corrode, partly because although that can be stopped it would be at the expense of the pipes thermal conductivity which is already very bad for air (so you would need too many pipes to make it worth while).
    My method is unique only as a redesign; the components it uses are about 99% already in existence. That’s why I'm so confident.
    Anyway thanks for your help; I intend to post my invention-process on this forum, under this Mechanical section, once I’ve got so far as proving beyond doubt I am the inventor.

    Thanks again.
    Yours, Alex.
     
  11. Jan 29, 2008 #10

    russ_watters

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    Yeah, I'm wondering if he's re-inventing the sterling engine...

    http://en.wikipedia.org/wiki/Sterling_engine
     
  12. Jan 29, 2008 #11

    Mech_Engineer

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    Sounds like you're trying to add a regeneration stage to a Brayton Cycle machine. If you look up regeneration in a thermodynamics text book you will be able to calculate the actual efficiency savings that would be possible using a regen cycle on a Brayton Cycle machine.

    Good luck!
     
  13. Jan 29, 2008 #12
    Are you familiar with the Hitachi heat recovery system for power plants? You might get some good info there.
     
  14. Jan 30, 2008 #13

    Mech_Engineer

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    Sounds like a stand alone regeneration cycle that is meant to be used on a steam powerplant, which would be modeled using the Rankine Cycle. That, or a regeneration cycle for a gas turbine plant which would of course be Brayton Cycle.

    Single or even dual stage regeneration cycles can recover a significant amount waste heat that would be lost to the environment, and as such they result in a significant efficiency savings as well.
     
    Last edited: Jan 30, 2008
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