Did I get double 50's?

[Luis H]

I have 14 steel 72 tanks that I have been taking some precise measurements.

Wall thickness measurements were taken using a precision ultrasound measuring device. The ultrasound probe was calibrated before each set of readings using a cut off (condemned) steel 72 tank. The cut off tank wall thickness could be confirm with a caliper and provided a sample of the exact same material to calibrate the ultra sound equipment. Measurements are good to about 2/1000 of an inch.

I measured the actual internal volume by using water at room temperature and precisely measuring the weight of the water. I took into account the reduced volume from the valve threads.

I am only showing some of the data below to save space. The “Tank #” is only the number I assign and painted on all my tanks to keep track of them. I have full records (in a spread sheet) with all the tank information, including serial number and all the tank markings (including all the hydro test markings).

I also have records of all the hydro test data, etc.

I am only including seven of my tanks at this point (plus a friends Walter Kidde tank). I still need to take precise volume data on the other tanks. I do have wall thickness for all of them. I need wall thickness and volume data to calculate REE numbers for that specific tank.

At this point I have complete data for four Norris tanks, three PST tanks, and a friends Walter Kidde tank (I don’t own any Walter Kidde).

The data is shown as follows:
Tank# / manufacturer >> average wall thickness / tank empty weight (with boot and valve) / tank actual volume at 2475 psi / actual REE nunber

1 / Norris >> 0.177 inches / 30.26 Lbs / 72.2 cu ft of air (@ STP) / REE = 61.3
2 / PST >>> 0.171 inches / 28.88 Lbs / 70.5 cu ft of air (@ STP) / REE = 58.6
4 / PST >>> 0.181 inches / 30.11 Lbs / 70.8 cu ft of air (@ STP) / REE = 61.5
5 / Norris >> 0.179 inches / 29.73 Lbs / 71.4 cu ft of air (@ STP) / REE = 60.7
6 / PST >>> 0.173 inches / 30.43 Lbs / 70.9 cu ft of air (@ STP) / REE = 58.7
7 / Norris >> 0.178 inches / 31.01 Lbs / 70.3 cu ft of air (@ STP) / REE = 58.4
8 / Norris >> 0.175 inches / 29.73 Lbs / 71.4 cu ft of air (@ STP) / REE = 61.2
No# / WK >> 0.183 inches / 29.63 Lbs / 71.0 cu ft of air (@ STP) / REE = 63.6


The variation in volumes is not very much considering the manufacturing process used for these tanks. As a mater of fact I consider the variation in wall thickness to be fairly impressive. The wall thickness standard deviation for the data collected for each tank is also relatively small.


As it has been mentioned, tanks with a + stamp (DOT 3AA code) are always advertised as having the capacity with the extra 10 % overfill.

An interesting side note. I also measured two PST HP 80s. They are supposed to have 80 cu ft of air at 3442 psi… they actually have 85.3 cu ft and 85.0 cu ft.

Note: as expected, the actual REE numbers are all higher (or equal) to the published number by PST. As a mater of fact I believe the REE number on tank 7 came out low due to lack of precision on the hydro test data. I believe I had a tiny leak during hydro, but not bad enough to do it over. I will confirm the data in 5 years.


Note2: using the equalization method (to get volume) is good for a rough approximation since you may start with an unknown quantity. You don’t really know for sure the donating tank actual capacity.

Also even a larger error will probably come from the pressure gauge measurement. Most pressure gauges are not very precise. I have being wanting to buy a digital pressure gauge (for gas mixing, miscellaneous “science projects”, etc.)


I hope this helps.

 

[John C. Ratliff]

For those of you who haven't seen The New Science of Skin and Scuba Diving, New Revised Edition by the Council for National Co-operation in Aquatics, these calculations come from Charles Law

PV = knT, or Pressure times Volume equals a constant (actually two constants) times the temperature.

The combined statement also represents Boyle's Law, which says that at a constant temperature, the volume of a gas varies inversely with pressure. These relationships may be easily checked with a toy balloon filled with air and dipped in water at varying temperatures or taken from the surface to the bottom of the deep end of a pool (constant temperature).

Since the above equations are scientific laws, they must hold for all conditions. Therefore, if a given amount of gas under a specified set of conditions P1, V1 and T1, is moved to a new environment, P2, V2, and T2, the first equation may be divided by the second (since both sides of each equation are equal) and the constant cancels out.

P1V1/P2V2 = kT1/kT2

For small changes in absolute temperatures (such as those encountered in ordinary diving) the ratio of T1 / T2 will be nearly unity. Since the error introduced by assuming the ratio to be 1 for small temperature changes is again within the limit of error in reading the usual type of diving pressure gauge, the equation can be rewritten in this form:

P1V1/P2V2 = 1 or P1V1 = P2/V2 (I wrote in the margin years ago, "Boyle's Law")

This above equation is known as the General Gas Law, as it combines Boyle's Law and Charles' Law (page 1-11 of the NOAA Diving Manual, Diving for Science and Technology, 1974 (I think; no copyright statement).

It is from this simple relationship that we get the ability to calculate the volume of a tank or set of tanks that is unknown if we start with two knows (pressure and volume) and measure the third (the pressure of the other tank. We can then solve for the volume of that second tank system. But we need to start with the correct pressure, and especially the correct volume, of the known tank so that we can calculate the volume of the second tank. It looks like this when we divide both sides of the equation by V2:

P1V1/P2 = V2

Now, realize that V2 is the total volume of both vessels, the known and the unknown. So if we know the pressure of the AL 80 is 3000 psig, and the volume is actually 77.4 cubic feet (according to Duckbill above), then all we need is to find the equalized pressure when the two systems are allowed to come to equilibrium, and the temperature is allowed to be the same. If that equalization pressure is 1500 psig, then the total volume will be calculated to be 154.8 cubic feet. If we then subtract 77.4 cubic feet from that, we get another cylinder at 77.4 cubic feet. If on equalization the pressure is 1000 psig, then the total volume is 232.2 cubic feet. Subtract the 77.4, and we get a set of twins at 154.8 cubic feet. If the equalization pressure is 2000 psig, then we get a total volume of 116.1 cubic feet, minus the original 77.4 cubic feet, and we get 38.7 cubic feet.

This is the same as what was described above by Duckbill and BJHawthorne, but this give a bit better theoretical description, again from a vintage text (first copyrighted in 1957, and my third edition is 1970).

SeaRat

PS--I wrote this before seeing Luis' post above. Thanks Luis, that is great info too.

 

[duckbill]

P1V1/P2 = V2

Now, realize that V2 is the total volume of both vessels, the known and the unknown. So if we know the pressure of the AL 80 is 3000 psig, and the volume is actually 77.4 cubic feet (according to Duckbill above), then all we need is to find the equalized pressure when the two systems are allowed to come to equilibrium, and the temperature is allowed to be the same. If that equalization pressure is 1500 psig, then the total volume will be calculated to be 154.8 cubic feet. If we then subtract 77.4 cubic feet from that, we get another cylinder at 77.4 cubic feet. If on equalization the pressure is 1000 psig, then the total volume is 232.2 cubic feet. Subtract the 77.4, and we get a set of twins at 154.8 cubic feet. If the equalization pressure is 2000 psig, then we get a total volume of 116.1 cubic feet, minus the original 77.4 cubic feet, and we get 38.7 cubic feet.

Yes, and, again, this works if the pressure rating of the unknown vessel is the same as the donor vessel. To calculate the capacity at dissimilar pressure raings, you would need to multiply the answer by the ratio of the two as I mentioned in a previous post.

 

[bjhawthorne]

Thanks for all the great information guys. The accuracy of the unknown volume calculation is very much a function of knowing the volume of the initial cylinder. I am most impressed with the technical excellence and information sharing here. My hat is off to you all.

 

[oldmossback]

All these forumulas are giving me a headache............I've some simple questions (simple multianswer would be appreciated)

Luis, do you have the numbers for wall thickness for the USD and Voit twin 38 cuft tanks? Mine are 3AA, chrome-moly.....rated 1800psi and 1880 psi plus 10%, respectively. The USD has a 3/4 neck and the Voit is 1/2 with the higher pressure.

I'd like to know the max test pressure and the margin between plus 10% and this test pressure.

If I do an air download to determine the volume of each, with a steel 71.2 at 2475psi.......which figure do I use.......the stamped psi on the tanks or the plus 10% rating.

I's asking this because I routinely fill my twins to 2200 or 2250 and let them cool back to 2150/2175 +- 15psi. These are my tanks, my risk, no one uses them but me.........still I'd like to know the margins..........and the actual est. volume.

For lake diving they give me all the air I need on a dive, mostly with 600-900 psi leftover.

 

[duckbill]

If I do an air download to determine the volume of each, with a steel 71.2 at 2475psi.......which figure do I use.......the stamped psi on the tanks or the plus 10% rating.

Mossback, you can use whatever pressure you want.

Try this. Plug in the values and calculate the innermost parentheses first:

P1 - Initial pressure reading of donor steel 72 (must be a basic steel 72 for this formula to work) when normalized at room temperature (Whatever your accurate gauge reads. Does not have to read 2475psi)

P2 - Pressure reading after equalization with ?cylinder and allowed to normalize to same room temperature as before.

R - Rated pressure of the ?cylinder (Including +10% if the original, factory hyro is "+")

Z - ...will be the approximate actual volume of your ?cylinder at the pressure you plugged in for "R"



R x ( (P1 - P2) x .0288 ) / P2 = Z

 

[John C. Ratliff]

Duckbill,

Please explain the physics behind needing to know the rated pressure of the cylinder. To me, that only applies in the calculation of the number of cubic feet of air that the cylinder holds at the rated pressure. That is the basis for figuring out the cubic feet that the cylinder being tested would hold, as it is a known quantity. If you don't start at the rated pressure, then the volume of the known cylinder's air in it would need to be multiplied by the ratio of the starting pressure to the rated pressure. That way, you can plug in the actual volume in the "V1" part of the equation.

Are we both talking about the same thing here, only saying it differently? I need to know as I have a new set of doubles I will be filling in this manner to make this calculation.

By the way, there is another potential problem, in that you don't ever want to overfill the cylinder being tested, even if it is inadvertently. Look at the rated pressure, and if you are starting out with a cylinder which is over that (say, you are filling a 2250 psi cylinder off a 3000 psi cylinder, and the former is fairly full) and don't fill it over that rating. From what I can tell, I can still make the calculations based upon the amount of psi lost/gained in each even if it is not equalized.

SeaRat

 

[duckbill]

Duckbill,

Please explain the physics behind needing to know the rated pressure of the cylinder. To me, that only applies in the calculation of the number of cubic feet of air that the cylinder holds at the rated pressure. That is the basis for figuring out the cubic feet that the cylinder being tested would hold, as it is a known quantity. If you don't start at the rated pressure, then the volume of the known cylinder's air in it would need to be multiplied by the ratio of the starting pressure to the rated pressure. That way, you can plug in the actual volume in the "V1" part of the equation.

John, when you mention "V1", I assume you are replying to my post #23, not my last post. I will have to think a bit more about it, but I think the problem is due to the fact that the volumes "V1" and "V2" relate to the physical volume of the container itself (physically measurable cubic inches, for example), not the usable volume of gas under compression. If you know the physical volume of the aluminum 80 (I think it should be about 652 cu.in.), then yes, you would subtract it from the new "volume" to find the volume of the unknown cylinder. But, we're talking physical volumes. You would then still have to figure the volume of air held by that cylinder at a given pressure.

I tried plugging in values using physical cubic inches, then converting to volume of gas at pressure and it works, but I don't have any more time to get into it now. I believe my explanation above is the reason the simple pressure/volume relationship is only for the physical volume(s), not the potential volume of the gases.

By the way, for an aluminum 80 donor cylinder, the equation would be:
R x ( (P1 - P2) x .0258 ) / P2 = Z
.....following the pattern I gave in my last post.

By the way, there is another potential problem, in that you don't ever want to overfill the cylinder being tested, even if it is inadvertently. Look at the rated pressure, and if you are starting out with a cylinder which is over that (say, you are filling a 2250 psi cylinder off a 3000 psi cylinder, and the former is fairly full) and don't fill it over that rating. From what I can tell, I can still make the calculations based upon the amount of psi lost/gained in each even if it is not equalized.

Unless the unknown cylinder is smaller than a typical scuba cylinder (Say, less than 38CF) and/or has a lower service pressure than the usual least seen on scuba cylinders (typically 1800psi), there is very little chance that a steel 72 is going to overpressurize it. One can always keep track of the pressure as it is equalizing and stop if the pressure exceeds the vessel's rating. Then, just empty the unknown cylinder again and redo the procedure starting with a lower pressure in the donor 72.

Even if the unknown cylinder happens to be a single 1800psi 38CF cylinder, it will equalize to just 1429psi using a steel 72 @ 2475psi as the donor, well below the 1800psi rating. No worries.

I admit I'm a bit rusty on math, so if anyone with good math skills finds fault I'd appreciate correction.

 

[elmer fudd]

You should be able to calculate the volume even if you don't drain the tank first.

If you know the starting pressure, the ending pressure and how many cubic feet of air you added, you can figure out what percentage of air is added and from that the overall volume of the tank.

 

[duckbill]

You should be able to calculate the volume even if you don't drain the tank first.

If you know the starting pressure, the ending pressure and how many cubic feet of air you added, you can figure out what percentage of air is added and from that the overall volume of the tank.

You can calculate the unknown cylinder's rated volume with air in it beforehand, as you say, but I'm pretty sure it would require a simultaneous equation which is much more complicated than what we have been discussing so far.