What does internal volume have to do with it? If you know the rated capacity and pressue, that's all you need to know to figure capacity at different pressure.
120/3500 = X/3200
Phil
HP tank Valves
- Thread starter sylvester
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What does internal volume have to do with it? If you know the rated capacity and pressue, that's all you need to know to figure capacity at different pressure.
120/3500 = X/3200
Phil
OP
sylvester
Guest
That was my next question, but you beat me to the punch. Based on what I am reading here with some being assumptions I understand that I will gain very little volume 4cf +/- and the real positive part of a swap is only my wgt saving differance between a lp104 and a hp120. Does any one know if the OMS lp and hp tanks are threaded the same?
lp are 3/4", hp are 7/8" far as I know. You can't swap valves between HP and LP steel tanks.
Phil
OP
sylvester
Guest
NT
Stone
Contributor
Your equation is an estimate that works great for scuba tanks (small internal volumes), and it is especially accurate when the actual tank pressure is near the working pressure.
The equation that uses the internal volume of the tank takes into account that, at 0 psig, there is still 1 atm of gas in the tank. Your equation says that, at 0 psig, there is a vacuum in the tank (0 cf of gas).
At a tank pressure of 3500 psig (assuming a service pressure of 3500 psig) your equation and the equation that takes internal volume into account result in exactly the same answer. As the tank pressure heads toward 0 psig, the error between the two equations increases until it reaches the internal volume in the tank. In the case of an HP 120, that means an error of 14.1 liters.
The equation that uses internal volume is only an estimate as well, because pressure vs volume is not really linear.
Ok lets look at what you want to do. If you know what kind of tank you are getting we can help you. If you buy a HP tank from, P.S.,faber, they come with a 3/4 neck same as LP If you get one from genesis they come with a small neck.And I do think they are the only one's that do. Ok lets see if I can get this right if you have a 300bar Din valve you can't put a plug into it because they made it fater so the yoke we not fit over the valve. If it is not 300 bar you can adapt it to yoke with a plug. Now here is how to find cf-psi. 120/ by 3500=.034x100=3.4cf/100psi=3500x3.4=119cf.
You can do this with all tanks to find the cf at the psi you want to use. Hope this helps
You can do this with all tanks to find the cf at the psi you want to use. Hope this helps
oxyhacker
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An "estimate" was all he needed. Why all the quibbles?
And anyhow still don't need volume even if you want to take into account PSIA/absolute pressure. Just add 15 psi to the starting point and finishing point and do the same math, to compensate for the fact that we don't really suck tanks dry, and that we measure fills by PSIG not PSIA. So:
(3215/3515) X 120=109.76 (rather than the 109.71 we get using PSIG). That's a difference of .05 psi which really isn't significant enough to demote such calculations to "estimates".
Using absolute pressure at both ends means that the alleged error is taken into account at both ends and doesn't grow gradually as the tank approaches empty, as you describe. Actually it doesn't grow as you describe even if one uses PSIG because the same extra 14 liters is present in the full tank as the empty one.
And anyhow still don't need volume even if you want to take into account PSIA/absolute pressure. Just add 15 psi to the starting point and finishing point and do the same math, to compensate for the fact that we don't really suck tanks dry, and that we measure fills by PSIG not PSIA. So:
(3215/3515) X 120=109.76 (rather than the 109.71 we get using PSIG). That's a difference of .05 psi which really isn't significant enough to demote such calculations to "estimates".
Using absolute pressure at both ends means that the alleged error is taken into account at both ends and doesn't grow gradually as the tank approaches empty, as you describe. Actually it doesn't grow as you describe even if one uses PSIG because the same extra 14 liters is present in the full tank as the empty one.
Stone
Contributor
Oxyhacker,
I don't see any quibbling. Phil asked me "what internal volume had to do with it" and I simply explained that my spreadsheet just happens to account for the volume of gas in an "empty" tank.
I do the math for fun, and when you use a spreadsheet, accounting for the tank volume doesn't slow down the calculation.
There is no doubt Phil's equation says there is no gas in the tank at 0 psig, when there is indeed .49 cf (in an HP120, at 70% humidity, at 65 F, at sea level, on the 23rd parallel, at noon, on a Sunday).
I don't see any quibbling. Phil asked me "what internal volume had to do with it" and I simply explained that my spreadsheet just happens to account for the volume of gas in an "empty" tank.
I do the math for fun, and when you use a spreadsheet, accounting for the tank volume doesn't slow down the calculation.
There is no doubt Phil's equation says there is no gas in the tank at 0 psig, when there is indeed .49 cf (in an HP120, at 70% humidity, at 65 F, at sea level, on the 23rd parallel, at noon, on a Sunday).
Yes but per the Pressed Steel WebPage the tank rating is not in psig it is in psi. So 3500 psi = 3514.7 psi (in an HP120, at 70% humidity, at 65 F, at sea level, on the 23rd parallel, at noon, on a Sunday, in this dimension).
If his equation is based, as the tank specification is, on psi rather than psig, his equation is correct.
Mike
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