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NotaPublicado: 31 Jul 2012 18:30 

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Halcyon Days
(fwd) bmk@dsi.bc.ca
(Barrie Kovish) 26 Feb 1997 16:08:01
-0500

>pH wrote: >>So consequently the loop volume has to be at
least 15 times the volume of >O2 consumed? Let's say I'm George IV and
actually interested in supplying >O2 to my brain and not just my superbly
conditioned O2 hogging muscles. I >manage to consume O2 at 3L min,
breathing 30 times a minute. That's 100cc >of O2 per breath. The loop's
volume must have two parts to it - the part >that you can breathe, i.e.,
suck down and the part that you can't, the >machine's deadspace. (I know
that this is not engineering hyperbole, bu >bear with me -)
>So for the loop to deliver me 100 cc of O2 (say 30% nitrox) the
loop >*breathable* volume is 15 x 100 x 3 ” .5 L ? Seems to me that you'd
have >to have another 3-4 L for the scrubber/tubing deadspace to total out
at >about 8 L in volume. Surely that's one hell of a liftbag at 100 fsw?
>What am I not understanding here? I actually don't think that
a loop volume of 8 L is particularly large, 4.5 l tidal volume is probably a
bit high but not impossible. An 8 l loop volume may actually be a little
low. Anyway in the conditions above you consume ..1 L (at STP) of O2 per
respiration. I claim that the PO2 of the expired gas will be 1/K (1/15 to
1/30) of an ATA less than the PO2 of the inspired gas. Given these
conditions you want to calculate the tidal volume Vt of the respiration, OK?
I believe that will be given as Vt ?dV or 15*.1 _ ?t _ “ 0*.1 or 1.5
Liters _ Vt _ 3.0 Liters Which I think are pretty standard tidal
volumes when working hard. To justify this formula here is a bit of a
derivation: Start with the ideal gas law PV ?RT. We've specified the O2
consumption in liters ( at STP ) which is a bit inconvenient. The volume of
O2 consumed is actually dV/P ie the volume consumed will decrease with depth.
To convert to change in molecules we have dn ?V/RT. The change
in PO2 is given (assuming constant Vt see * below) by dPVt ?nRT
or Vt ?nRT/dP substituting in dn we get Vt
?V/dP dP is ( I claim ) 1/K and hence Vt
?dV Wew**! Barrie*** *Note that this whole analysis is
thrown out a bit by the fact that the volume is not totally consant. The O2
consumed is not nessesarily totally replaced with CO2. However the ratio of
CO2 produced to O2 consumed is ,I believe, somewhere between
0.8-1.0. **None of this of course proves my claim the dP is 1/K! However
there is a simple experiment to check this out ( at least on the surface ).
Simply breath over an O2 sensor. Note 1 confounding factor is that the
PCO2 will rise over the course of the breath ( mostly I think during the
first 1/3 ) and hence PO2 will drop. ***Good idea and interesting
question to test the validity of the parameter K by looking at the implied
tidal volumes.


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