Ean 18 ???

[Divrdug]

I no what he dun, he just breethd off the O2 part of the tank. Now dont go makin this tuffrrr thain it shud be!
The divn hillbily

 

[Thalassamania]

Oh ho, I think steel could rust enough. Let’s see what pours out.

 

[H2Andy]

which would indicate a problem with my analyzer).

i doubt it, since it correctly analyzed two tanks with known O2 %

 

[loosebits]

Ok, lets do the math (I'm bored).. can a steel tank rust enough to get the O2 down from 21% to 18% assuming a 3000 psi fill in a 12 L tank (AL80 sized)?

Lets assume that a steel tank is made up of 100% iron. Rusting is as follows:

2 Fe + 1.5 O2 -> Fe2O3.

Now we need to know how much O2 needs to be consumed to go from 21% to 18%:

(.21 - .18) * 3000 psi * 12 L / 14.7 psi = 74.5 L of O2 = 3.28 mol (22.4 L/mol of any ideal gas at STP)

Ok, so with 3.28 mol of O2 available, how much Fe will it consume:

3.28 mol / 1.5 * 2 = 4.37 mol of Fe.

Now I am curious how much the tank would have to rust to use 4.37 moles of iron (how thick would the rust have to be).

4.37 mol * 55.85 g/mol / 7.86 g/cm^3 = 31.07 mL

Ok, now how deep into the tank would it have to rust to represent 31 mL of Fe?

Lets assume the tank is a cylinder.. the volume of a rusting layer is give by:

volume = pi * h * (r^2 - (r - t)^2)

where r is the radius of the tank, h is the height and t is the thickness of the layer of iron that must rust.

So, we want to solve for t. Since we are only talking about 200 or so g of iron here and the weight of a tank is about a hundred times more, we can assume t is very small compared to r so the rearranged equation, V/(pi * h) = 2rt - t^2 simplifies to

t =~ V/(2r * pi * h)

(I removed the t^2 term and solved for t)

Using a 25" tall tank, 7.25" in diameter (3.625" radius), it seems we only need 0.0033 inches of steel to rust. That certainly seems possible that .0033" of steel could be exposed to O2 during the course of rusting.

Personally I find this result surprising and might have screwed up the math somewhere.. I wouldn't think a tank could rust enough to make a significant difference in the FO2 at 3000 psi.

Here's the google calc I used (google calculator is amazing): http://www.google.com/search?num=50&hl=en&safe=off&rls=GGGL%2CGGGL%3A2006-27%2CGGGL%3Aen&q=%28%28.21+-+.18%29+*+3000+lb%2Fin%5E2+*+12+L%2F14.7+lb%2Fin%5E2+%2F+22.4+L%2Fmol+%2F+1.5+*+2+*+55.845+g%2Fmol+%2F+7.86+g%2Fcm%5E3%29+%2F+%282+*+3.625+inches+*+25+inches+*+PI%29+in+inches&btnG=Search

 

[DandyDon]

It's a steel tank. I'm going to visit the nitrox shop today to first see what their analyzer reads, then drain & viz (unless theirs reads fine, which would indicate a problem with my analyzer).

Steel tank, rust, oxidation...?

What about this question...??

And you said "from a shop that only fills air" - did you have scuba air not cleaned to Nitrox standards added to a O2 cleaned tank...?

 

[Shallow Draft]

I think you are in error..... as the decimal point in line 2 is in the wrong place
This is clearly a case of too much time on dry land!!!

Ok, lets do the math (I'm bored).. can a steel tank rust enough to get the O2 down from 21% to 18% assuming a 3000 psi fill in a 12 L tank (AL80 sized)?

Lets assume that a steel tank is made up of 100% iron. Rusting is as follows:

2 Fe + 1.5 O2 -> Fe2O3.

Now we need to know how much O2 needs to be consumed to go from 21% to 18%:

(.21 - .18) * 3000 psi * 12 L / 14.7 psi = 74.5 L of O2 = 3.28 mol (22.4 L/mol of any ideal gas at STP)

Ok, so with 3.28 mol of O2 available, how much Fe will it consume:

3.28 mol / 1.5 * 2 = 4.37 mol of Fe.

Now I am curious how much the tank would have to rust to use 4.37 moles of iron (how thick would the rust have to be).

4.37 mol * 55.85 g/mol / 7.86 g/cm^3 = 31.07 mL

Ok, now how deep into the tank would it have to rust to represent 31 mL of Fe?

Lets assume the tank is a cylinder.. the volume of a rusting layer is give by:

volume = pi * h * (r^2 - (r - t)^2)

where r is the radius of the tank, h is the height and t is the thickness of the layer of iron that must rust.

So, we want to solve for t. Since we are only talking about 200 or so g of iron here and the weight of a tank is about a hundred times more, we can assume t is very small compared to r so the rearranged equation, V/(pi * h) = 2rt - t^2 simplifies to

t =~ V/(2r * pi * h)

(I removed the t^2 term and solved for t)

Using a 25" tall tank, 7.25" in diameter (3.625" radius), it seems we only need 0.0033 inches of steel to rust. That certainly seems possible that .0033" of steel could be exposed to O2 during the course of rusting.

Personally I find this result surprising and might have screwed up the math somewhere.. I wouldn't think a tank could rust enough to make a significant difference in the FO2 at 3000 psi.

Here's the google calc I used (google calculator is amazing): http://www.google.com/search?num=50&hl=en&safe=off&rls=GGGL%2CGGGL%3A2006-27%2CGGGL%3Aen&q=%28%28.21+-+.18%29+*+3000+lb%2Fin%5E2+*+12+L%2F14.7+lb%2Fin%5E2+%2F+22.4+L%2Fmol+%2F+1.5+*+2+*+55.845+g%2Fmol+%2F+7.86+g%2Fcm%5E3%29+%2F+%282+*+3.625+inches+*+25+inches+*+PI%29+in+inches&btnG=Search

 

[loosebits]

Hmm, I must be blind, don't see it but I agree, too much time on dry land.

 

[yak]

Weeeelllllll, case closed, but mystery not solved.

Brought the tank and the analyzer to the shop. They analyzed a tank of 36% with theirs and with mine and both read the same. Then they read my tank and it read 21% with theirs AND MINE!!! So I look like an idiot in the shop (and on this board):shakehead

Could it take a few days for the sensor to stabilize after unsealing it?

 

[DandyDon]

Operator error...!! I heard of it.

 

[RichKirby]

Operator error...!! I heard of it.

Well, that only took five pages of posts. This is funny ONLY because it wasn't me. Sorry, Yak, no offense meant, just poking a little fun.