Tank Valve Freeze Up

[Graeme Tolton]

Go get a can of compressed air like you use for electrical components and see how long you can hold the trigger down before it starts getting extremely cold and freezing up

My question is the OP said something about being upside down, if there was any moisture in the tank this would have put it near the valve assembly, add a free flow situation or high rate of flow could this have caused this? Interesting they didn't notice a leak while diving though.

These cans are not compressed air, they are usually filled with some sort of liquid refrigerant. The liquid expanding into a gas is what causes the extreme cold from these cans. The change of state is where the rapid temperature drop is from. If you spray them inverted, you will see that they are liquid, like butane or propane. The expanding high pressure air will get cold, but should never get that cold when it is so warm.

I think there must be more to this story than we are seeing...

 

[ClayJar]

Okay, I had to pull a little physics out for this one...

First let's figure adiabatic expansion of tank air to the outside:

PV^γ = K​

K is a constant, as is γ, which is 1.4 for air.

P₁V₁^γ = P₂V₂^γ
P₁V₁^1.4 = P₂V₂^1.4​

Now, let's set a few values:

P₁ = 1500 psi
P₂ = 33 psi (at 40 fsw)​

Which we can now plug in, rearrange, and simplify:

(1500 psi)V₁^1.4 = (33 psi)V₂^1.4

(1500 psi / 33 psi) = V₂^1.4/V₁^1.4

(1500 psi / 33 psi)^(1/1.4) = V₂/V₁

15.3 = V₂/V₁

(15.3)V₁ = V₂​

Taking the ideal gas law, n (number of moles, i.e. amount of stuff) and R are constants, so we can take it, make the General Gas Law, plug in our values, and simplify:

PV = nRT
P₁V₁/T₁ = nR = P₂V₂/T₂

(1500 psi)V₁/T₁ = (33 psi)((15.3)V₁)/T₂

(1500 psi)/T₁ = (33 psi)(15.3)/T₂

((1500 psi)/((33 psi)(15.3))) = T₁/T₂ = 2.97

T₁/2.97 = T₂​

Rounding 85 degrees F to 30 degrees C, we find our temperature:

T₁ = 30 degrees C = 303 K
T₂ = T₁/2.97 = (303 K)/2.97

T₂ = 102 K​

So, now let's assume that adiabatically-expanded air absorbs heat from the water and only the water in an amount sufficient to make its final temperature in equilibrium with the ice. (We'll use fresh water for convenience. With the approximations we have to use, it's not significant to this calculation.)

T_adiabatic = 102 K
T_final = 0 degres C = 273 K
C_{p, air} = 1 J/(g * K) (heat capacity of air)

ΔH_air = (T_final - T_adiabatic) * C_{p, air}

ΔH_air = (273 K - 102 K) * (1 J / (g * K))

ΔH_air = 171 J/g​

So, we now know how much heat is required for the expanding air. We can now see how much heat we must absorb to create ice from our 30 degree Celsius water. This is done in two halves. The first half is cooling it to the freezing point, and the second half is actually freezing it (the "heat of fusion" as it's called):

ΔH_water = ΔH_{water, cooling to freezing} + ΔH_{water, freezing}

ΔH_{water, cooling to freezing} = (T_hot - T_cold) * C_{p, water}
C_{p, water} = 1 J / (g * K) (yeah, it's about the same)

ΔH_{water, cooling to freezing} = (303 K - 273 K) * (1 J / (g * K)) = (30 K) * (1 J / (g * K))
ΔH_{water, cooling to freezing} = 30 J/g

ΔH_{water, freezing} = 334 J/g (you just look that up)

ΔH_water = ΔH_{water, cooling to freezing} + ΔH_{water, freezing}
ΔH_water = 30 J/g + 334 J/g
ΔH_water = 364 J/g​

So, we now have our two pretty values. Assuming that all the water to heat the expanding air comes from the water being frozen, we can set these equal. (In real life, there will be heat transfer through the tank, to moving water, from *you*, and on and on, but assuming it *all* comes from the water being frozen gives you a hard upper limit.) We have to multiply each times its respective mass to get the total heats Q, to set equal. Doing this, rearranging, and simplifying:

Q_air = m_airΔH_air = m_air(171 J/g_air)
Q_water = m_waterΔH_water = m_water(364 J/g_water)

Q_air = Q_water
m_airΔH_air = m_waterΔH_water
m_air(171 J/g_air) = m_water(364 J/g_water)

m_air/m_water = (364 J/g_water)/(171 J/g_air)

m_air/m_water = 2.13​

Sooo... you need 2.13 times as much mass of air as mass of water, but what does that mean? Well, if you assume that you're using an aluminum 80 that has six pounds of air inside when filled to 3000 psi, that comes out to about 0.91 grams per psi. Since we're already assuming no heat transfer through the tank and all that, we don't lose anything by going with round numbers, which gives us about 2 psi / gram of ice, i.e. 2 psi / cc of water frozen.

Now, all of this comes down to what, exactly? Well, assuming your ice cube tray uses the standard 1 fluid ounce per cube, i.e. about 30cc, you could freeze one cube's worth of ice with 60 psi of free flowing, *assuming* you didn't lose any of that "cold" to or through anything else.

Now, all that said, I can't imagine any situation where you could actually freeze water to your valve and first stage with anywhere near even an order of magnitude of that rate, but it's apparently not *completely* impossible. Absolutely implausible, but a quick basic physics check doesn't actually rule it out (which I found interesting).

Anyway, I guess that's about it for today's Fun With Physics post. If anyone wants to check my work, feel free. I'm *far* too tired to think I didn't miss something. :biggrin:

 

[RJP]

If anyone wants to check my work, feel free.

Nah, I think I'm good with what you've done.

 

[RJP]

As far as the incident ... There was definitely enough ice to "scoop" out of the tank valve mouth and the opening of the first-stage,

Ice INSIDE the valve would have to mean water INSIDE the tank, no?

 

[katdiver]

Sounds feasable

 

[DandyDon]

Ice INSIDE the valve would have to mean water INSIDE the tank, no?

Where else could it come from...?

The Op kept saying "bad tank," didn't he? A problem they've had before? I don't know enough about scuba air compressors to remember how they prevent water from getting inside a tank, but one commonly get precipitation inside shop compressor tanks which has to be drained periodically. Scuba air compressors operating in hight air humidity are required to prevent this of course, but how do you know about the quality of air in your tank? When asked if I could tour their compressor room and see their quarterly test records, some Ops have been proud, some have been insulted - and unable to produce test records. In both cases, I smell, taste, and analyze for CO tho.

If you accept that their compressor allow water vapor in, I guess you'd expect other problems.