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zepp Group Title

I have a question about breathing gas underwater. Here goes: As we go down, the quantity of gas breathed increases (because the pressure goes up) So for a diver that's going down at a speed of 18m/min until he reaches 60 meters underwater (It would take him 3 minutes and 20 seconds to do so). How would I calculate the amount of gas he's breathing during this 3 minutes and 20 seconds, knowing that the atmospheric pressure is 100kPa and the water's temperature is about 20C and that he's breathing rhythm is about 18L of gas per minute?

  • 2 years ago
  • 2 years ago

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  1. zepp Group Title
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    I tried to calculate the average amount of gas breathed by using the quantity of air breathed every minute before the dive and the quantity of air breathed every minute 60 meters underwater. At 0m, I could use the perfect gas formula, PV=nRT, where P = 100kPa; V = 18L; n = ?; T = 293K\[n = (\frac{100*18}{8.314*293})=0.73mol\]At 60m, P = 700kPa; V=18L; n=?; T=293K\[n=\frac{700*18}{8.314*293}=5.17mol\]and calculate the average, is that a good way to do it? (\(Average =(5.17+0.73) /2=2.95mol/minute\\2.95mol/minute*3.33 minutes =3.94mol\))

    • 2 years ago
  2. zepp Group Title
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    @TuringTest

    • 2 years ago
  3. zepp Group Title
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    Do you know if my work is correct for this kind of situation?

    • 2 years ago
  4. TuringTest Group Title
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    if air intake is a function of depth you would need an integral to do it right

    • 2 years ago
  5. zepp Group Title
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    Well, intake is a function of pressure which is a function of depth

    • 2 years ago
  6. Carl_Pham Group Title
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    I don't think your assumptions are correct. I would expect both respiration volume and frequency to decrease as the pressure of O2 increased. The urge to breath is driven by the accmulation of CO2 in the blood. Therefore I would suspect you do not breathe a fixed volume, but rather a fixed number of moles of O2 per minute. I would expect THAT to stay constant as the diver descended, so that the actual volume of O2 consumed at the higher pressure would steadily decrease.

    • 2 years ago
  7. zepp Group Title
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    Volume is related to the quantity of air, greater pressure doesn't mean taking little breathing. that's what I was told :/

    • 2 years ago
  8. zepp Group Title
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    because the first way I did was to calculate with a fixed quantity of gas

    • 2 years ago
  9. TuringTest Group Title
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    I honestly don't know about that, but IF air intake is a function of pressure, which is a function of depth, then you need to integrate air intake as a function of pressure for the trip.

    • 2 years ago
  10. Carl_Pham Group Title
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    No, that's silly. It's well known that prebreathing pure O2 tremendously reduces the urge to breathe for a while afterward. But this has nothing to do with your method of solving the problem -- it was just a throught I had.

    • 2 years ago
  11. zepp Group Title
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    Pressure is calculate with the formula P = pgh, where h = height, g = gravitational acceleration and p is the density of the medium we are going into. So \[\large \int_{0}^{60}(pgh)dh\]?

    • 2 years ago
  12. TuringTest Group Title
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    you need a term for air intake A (per unit time) as a function of pressure A(P(t))dt... I forget hydrostatics :P

    • 2 years ago
  13. zepp Group Title
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    @Carl_Pham, the diver is breathing normal air during his trip, but as he goes down, the partial pressure of O2 will go up, so he would breath more oxygen than he would at the surface?

    • 2 years ago
  14. zepp Group Title
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    Ah, we integrate wrt to time?

    • 2 years ago
  15. zepp Group Title
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    Oh, ok ok I see :P

    • 2 years ago
  16. TuringTest Group Title
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    or you could make pressure a function of depth A(P(h))dt

    • 2 years ago
  17. TuringTest Group Title
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    I would integrate wrt if your function is air intake per unit time I suppose you could do air intake per meter dived, but that seems like a tricky function

    • 2 years ago
  18. TuringTest Group Title
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    If air intake is a function of depth then how fast you are moving will affect where you are at a given moment, and how much air you have taken in up to that point

    • 2 years ago
  19. zepp Group Title
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    He's breathing at a rate of 18m/min, so integrating wrt time would be more appropriate I think, so I would have to compute a formula for pressure wrt time, is that correct?

    • 2 years ago
  20. TuringTest Group Title
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    you could do that if you take the divers path as fixed I guess, but pressure could be wrt depth without too much trouble

    • 2 years ago
  21. TuringTest Group Title
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    something like\[dP=\rho gdh\]\[dA(P)=f(\rho gh)\rho gdh\]A is some function of pressure f(P)

    • 2 years ago
  22. TuringTest Group Title
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    again, not really sure... haven't done fluid mechanics in forever it would be nice to assume air intake is directly proportional to pressure

    • 2 years ago
  23. zepp Group Title
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    Air intake.. that would be his breathing speed?

    • 2 years ago
  24. TuringTest Group Title
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    then you could have\[dA\propto dP=k\rho gdh\]

    • 2 years ago
  25. TuringTest Group Title
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    air intake as a rate I was thinking, but then you have to find the total air intake which would require integration wrt time, so... you would have to make h a function of t and integrate wrt time in the end anyway I suppose

    • 2 years ago
  26. zepp Group Title
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    Alright, I'll try it out, thanks!

    • 2 years ago
  27. TuringTest Group Title
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    Good luck. Thinking about it makes me remember why I didn't like fluid mechanics all that much.

    • 2 years ago
  28. zepp Group Title
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    In case of failure, I'll call you :P

    • 2 years ago
  29. TuringTest Group Title
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    \[A\propto P(t)=k\rho gh(t)\]\[dA=k\rho gh'(t)dt=k\rho gvdt\]something like that for starters I think. See ya!

    • 2 years ago
  30. TuringTest Group Title
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    the above if pressure is directly proportional to depth

    • 2 years ago
  31. TuringTest Group Title
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    I mean air intake propto depth

    • 2 years ago
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