SCBA tanks and heat?

They can blow a burst disc if they are filled to capacity in a cool house for example and then put in a car, left out in the heat on a 110 degree day and the temperature rises a lot. If there isn’t enough room to allow for expansion, the burst disc could blow. I’m sure it has happened but I’ve never seen it.

Stoti
 
So then it sounds like I could in large part avoid that problem by filling the tank when it's hot outside in the first place, right? Or at least not take the tank outside when it's full. That's doable -- I'm not going to be able to run my compressor in the house anyway.

By the way, how do you go about repairing a blown burst disc? Is that something I should even be trying to do by myself?
 
For simplicity and to account for the most basic affects, we can use the ideal gas law which is as follows:
PV=nRT

where:
P=pressure
V=volume
n=number of moles
R=gas constant
T=temperature

The variables V, n, and R are constant for a cylinder of air that just changes temperature with the same amount of air in it. Therefore, we can say that P is linearly proportional to T (and vice-versa). When T increases, P increases linearly, and when T decreases, P decreases linearly.

You can also think of it this way: since V, n, and R are constant, we can re-arrange the equation so that P and T are on one side and V, n, and R are on the other. This gives you:

P/T = nR/V

Let's say you fill your tank at 70 deg F which results in a pressure of 4500 psi. But then you move it to a hot car that causes the tank to reach 120 deg F (not unheard of in a hot car). Since nR/V (the right side of the equation above) is constant, you can say P1/T1 = P2/T2. You know P1, T1, and T2 so now you can solve for P2.

Note that I am converting deg F to K, since we need to use an absolute temperature scale in the ideal gas equation.

70 deg F = 294.3 K
120 deg F = 322K

P1/T1 = P2/T2
4500/294=P2/322
P2=322*(4500/294)
P2=4929 psi

So by leaving your 4500 psi cylinder filled at 70 deg F in a hot car and allowing the cylinder to reach 120 deg, the pressure will reach a little over 4900 psi! Cars can get even hotter than this, so this is definitely something to think about!

And yes, filling your tank initially at a hotter temperature while cause the pressure to not rise as much.

This is a good question to ask yourself if you ever have plans of packing up your gun and cylinder in the vehicle with plans to go to work all day and then go to the range afterwards. How hot is your cylinder and gun going to while you're at work?

-Clayton




Note: in the P1/T1=P2/T2 equation it is ok to use whatever pressure units you want since you are using the the same on both sides. The ideal gas law (PV=nRT) technically requires P to be in Pascals, V in cubic meters, and T in Kelvin when using the SI unit version of R=8.314 Pa⋅m3/(K⋅mol).



EDIT to correct my earlier error with the temperature units! (Thanks to those who pointed it out.)


 
For simplicity and to account for the most basic affects, we can use the ideal gas law which is as follows:
PV=nRT

where:
P=pressure
V=volume
n=number of moles
R=gas constant
T=temperature

The variables V, n, and R are constant for a cylinder of air that just changes temperature with the same amount of air in it. Therefore, we can say that P is linearly proportional to T (and vice-versa). When T increases, P increases linearly, and when T decreases, P decreases linearly.

You can also think of it this way: since V, n, and R are constant, we can re-arrange the equation so that P and T are on one side and V, n, and R are on the other. This gives you:

P/T = nR/V

Let's say you fill your tank at 70 deg F which results in a pressure of 4500 psi. But then you move it to a hot car that causes the tank to reach 120 deg F (not unheard of in a hot car). Since nR/V (the right side of the equation above) is constant, you can say P1/T1 = P2/T2. You know P1, T1, and T2 so now you can solve for P2.

P1/T1 = P2/T2
4500/70=P2/120
P2=120*(4500/70)
P2=7714.3 psi

So by leaving your 4500 psi cylinder filled at 70 deg F in a hot car and allowing the cylinder to reach 120 deg, the pressure will reach a little over 7700 psi!! Cars can get even hotter than this, so this is definitely something to think about!

And yes, filling your tank initially at a hotter temperature while cause the pressure to not rise as much.

This is a good question to ask yourself if you ever have plans of packing up your gun and cylinder in the vehicle with plans to go to work all day and then go to the range afterwards. How hot is your cylinder and gun going to while you're at work?

-Clayton




Note: in the P1/T1=P2/T2 equation it is ok to use whatever pressure or temperature units you want since you are using the the same on both sides. The ideal gas law (PV=nRT) technically requires P to be in Pascals, V in cubic meters, and T in Kelvin when using the SI unit version of R=8.314 Pa⋅m3/(K⋅mol).



Not questioning your math at all, but it indicates that at 120F the pressure would exceed the limit of the burst disc and the tank would vent itself. I feel like a piece of the puzzle may be missing here since firefighters, by design, use these exact same rigs for prolonged periods in areas which easily reach 300-400F+

Can't imagine they would use rigs that would be so likely to vent (and starve them of all of their air) at only 120F.

Edit: Also please note that I'm in no way recommending that anyone ever leave an SCBA in a hot car. I never do that, personally. Just working through the logic here for arguments sake.
 
For simplicity and to account for the most basic affects, we can use the ideal gas law which is as follows:
PV=nRT

where:
P=pressure
V=volume
n=number of moles
R=gas constant
T=temperature

The variables V, n, and R are constant for a cylinder of air that just changes temperature with the same amount of air in it. Therefore, we can say that P is linearly proportional to T (and vice-versa). When T increases, P increases linearly, and when T decreases, P decreases linearly.

You can also think of it this way: since V, n, and R are constant, we can re-arrange the equation so that P and T are on one side and V, n, and R are on the other. This gives you:

P/T = nR/V

Let's say you fill your tank at 70 deg F which results in a pressure of 4500 psi. But then you move it to a hot car that causes the tank to reach 120 deg F (not unheard of in a hot car). Since nR/V (the right side of the equation above) is constant, you can say P1/T1 = P2/T2. You know P1, T1, and T2 so now you can solve for P2.

P1/T1 = P2/T2
4500/70=P2/120
P2=120*(4500/70)
P2=7714.3 psi

So by leaving your 4500 psi cylinder filled at 70 deg F in a hot car and allowing the cylinder to reach 120 deg, the pressure will reach a little over 7700 psi!! Cars can get even hotter than this, so this is definitely something to think about!

And yes, filling your tank initially at a hotter temperature while cause the pressure to not rise as much.

This is a good question to ask yourself if you ever have plans of packing up your gun and cylinder in the vehicle with plans to go to work all day and then go to the range afterwards. How hot is your cylinder and gun going to while you're at work?

-Clayton




Note: in the P1/T1=P2/T2 equation it is ok to use whatever pressure or temperature units you want since you are using the the same on both sides. The ideal gas law (PV=nRT) technically requires P to be in Pascals, V in cubic meters, and T in Kelvin when using the SI unit version of R=8.314 Pa⋅m3/(K⋅mol).


You're basically correct in the calculation, but you can't use Fahrenheit, you have to use Kelvins. (F − 32) × 5/9 + 273.15

So 4500 at 70F is approx 4924 at 120F
 
The gas laws require absolute temperatures which is 273 + temperature in centigrade. Roughly the pressure at 120 F would be about 4900 psi or a 400 psi increase. I did not take time to convert F to C but estimated initial temp at 25 C and hot car at 50 C

This is the correct calculation. It does not jump up as much as you would think. Like kpaoletti said, the temps need to be in absolute temp form.

James
 
Well, I’d keep in mind that these SCBAs are primarily intended to be used inside of burning buildings, so...

That IS reassuring, now that you mention it! 😅


I do get a bit of a chuckle out of seeing us airgunners treating SCBAs like newborn infants and carrying them around in padded/reinforced bags and contraptions.

I used them for years in the service, and these things can be abused like you wouldn't believe. That's what they're made for. The hardest use in the hardest conditions imaginable.
 
There is enough safety margin factored in to allow for changes in pressure with temperature when regulations are drawn up. Meaning if your bottle was filled on a cold day and it is left in your hot car and the pressure rises by 300-400 PSI or more. It is still perfectly safe. A burst disc shouldn't blow off under these circumstances but sometimes might because they are not infallible. Like has been stated, these bottles are designed to be used in extreme conditions. The dive industry has pushed back against burst discs on their valves for years because they can't be trusted. It would be disastrous to have one fail 40 metres down. 
 
I only fill all my 4500psi rated tanks to 4000psi max no matter what. That would give ample safety Lee-way left in the trunk of a hot car. Reason only 4000psi I don't want to ruin any of my compressors because that last 500psi is THE Compressor killer. 

Same with my PCP guns only max 3000psi regardless of max rating less stress on seals and regulators don't don't seem to act as funky sometimes without warning compared to filling at only 3000psi. I am however mindful not to leave my 3000psi rated guns in the hot car for periods of time when filled to 3000psi.
 
For simplicity and to account for the most basic affects, we can use the ideal gas law which is as follows:
PV=nRT

where:
P=pressure
V=volume
n=number of moles
R=gas constant
T=temperature

The variables V, n, and R are constant for a cylinder of air that just changes temperature with the same amount of air in it. Therefore, we can say that P is linearly proportional to T (and vice-versa). When T increases, P increases linearly, and when T decreases, P decreases linearly.

You can also think of it this way: since V, n, and R are constant, we can re-arrange the equation so that P and T are on one side and V, n, and R are on the other. This gives you:

P/T = nR/V

Let's say you fill your tank at 70 deg F which results in a pressure of 4500 psi. But then you move it to a hot car that causes the tank to reach 120 deg F (not unheard of in a hot car). Since nR/V (the right side of the equation above) is constant, you can say P1/T1 = P2/T2. You know P1, T1, and T2 so now you can solve for P2.

P1/T1 = P2/T2
4500/70=P2/120
P2=120*(4500/70)
P2=7714.3 psi

So by leaving your 4500 psi cylinder filled at 70 deg F in a hot car and allowing the cylinder to reach 120 deg, the pressure will reach a little over 7700 psi!! Cars can get even hotter than this, so this is definitely something to think about!

And yes, filling your tank initially at a hotter temperature while cause the pressure to not rise as much.

This is a good question to ask yourself if you ever have plans of packing up your gun and cylinder in the vehicle with plans to go to work all day and then go to the range afterwards. How hot is your cylinder and gun going to while you're at work?

-Clayton




Note: in the P1/T1=P2/T2 equation it is ok to use whatever pressure or temperature units you want since you are using the the same on both sides. The ideal gas law (PV=nRT) technically requires P to be in Pascals, V in cubic meters, and T in Kelvin when using the SI unit version of R=8.314 Pa⋅m3/(K⋅mol).


You're basically correct in the calculation, but you can't use Fahrenheit, you have to use Kelvins. (F − 32) × 5/9 + 273.15

So 4500 at 70F is approx 4924 at 120F

This is the closest to being exact.

Welcome back!