# Containers Structurally Able to Hold Freezing Water?

1. Dec 9, 2004

What kinds of tanks are capable of resisting these types of pressures? (30,000 psi). I can't find any tanks that can. I would assume extremely thick steel could, but I can't find any specific tanks/containers in production that can. What naturally occuring rocks are able to not crack when water gets in its crevices and freezes?

2. Dec 9, 2004

### chroot

Staff Emeritus
What do you mean "resist?" Do you mean contain 30,000 psi without deforming? A standard soda bottle can "resist" the pressures, i.e. without releasing its contents, but it is typically deformed by the force.

- Warren

3. Dec 10, 2004

### alpha_wolf

Err... usually, you leave some air in the tank. This way, as the water freezes, it expands and compresses the air, so the pressure is kept far lower than 30000 psi (a bit more than atmospheric). Other than that, it's an issue of material and wall thickness..

4. Dec 11, 2004

Yeah, I'm talking about without deforming.

Are you sure a soda bottle could withstand that much without exploding?

5. Dec 12, 2004

### Staff: Mentor

Technically speaking, there is no such thing as "without deforming," since there is no such thing as a perfectly rigid material.

And yes, a plastic bottle certainly can withstand freezing (with deformation) - even an aluminum can will sometimes withstand it.

6. Dec 12, 2004

### Mk

Is this how amphoric ice is made? Freezing water without letting it expand?

7. Dec 12, 2004

Let us then define "without deforming" in this discussion as how much a scuba diving oxygen tank actually deforms when it is full.

8. Dec 12, 2004

I believe not allowing water to expand when it freezes would create a mixture of normal ice and "ice II."

9. Dec 14, 2004

### pack_rat2

Exactly. And even the slightest difference in pressure between the inside and outside of a tank will cause it to deform. The question is, will the deformation be permanent? If the stress is below the yield stress of the material, it will spring back to its original shape when the pressure is removed. If a tank has a wall thickness to diameter ratio of 1:20 or less, it's considered "thin-walled" and the following formula can be used to find the maximum stress (which is normal to the length of the cylinder, tangent the circumfrence) in the wall: stress = p*d/(2*t), where p is the pressure, d is the cylinder diameter, and t is the wall thickness. You need to know the dimensions of the tank to find the stress. Then you can check tables of material properties for yield stresses to find one that would work. If this were a real-world design problem, you'd also have to incorporate a safety factor, probably of 50%+ (it may be different).

10. Dec 15, 2004

### Mk

Do you guys get the idea or are you just nitpicking at his question?

When water is cooled below its normal freezing point, it normally freezes to form hexagonal ice. If it is very pure and cooled carefully, it may be supercooled to about -42°C. If water is cooled very rapidly then it forms a hyperquenched glassy water (HGW). HGW is formed by the spraying of a fine jet of water droplets into very cold liquefied gas (propane is commonly used) at around or below 80 K, involving cooling rates of greater than $$10^5 K$$ per $$s^-1$$.
A similar material is amorphous solid water (ASW or "low density glass"), formed from the slow deposition of water vapor, at less than 0.2 nanometers per second, on a very cold metal crystal surface below 120 K.
High density glassy water (HDG, $$1.1 g/cm^-3$$) is formed by vapor deposition at 10 K.
Rime is a type of ice usually formed by fog freezing on cold objects. It contains a high air:water ratio, making it appear white rather than transparent, and giving it a density about one quarter of that of Ice I.

11. Dec 15, 2004

So ehh... does anybody know of any specific types of containers that are able to withstand, without deforming (by the definition I gave earlier*), the expansion of freezing water? What thickness would a steel tank have to be?

*

12. Dec 15, 2004

### Staff: Mentor

Then yes, certainly you could build a tank to withstand 30,000 psi with that much deformation.
The question wasn't very specific...

A scuba tank is maybe half a centimeter thick and is rated for 3,000 psi. So, 10 times as strong would be 10 times as thick or 5cm (I may be oversimplifying, but I don't feel like running down the equations).

Carbon fiber would probably be my material of choice though: by volume, its properties are comparable to steel, while it's many times less dense. http://

Last edited by a moderator: Apr 21, 2017
13. Dec 15, 2004

### PerennialII

We have a couple of "environmental containers" at work which essentially are just steel tanks with stiffeners. Those are used e.g. in studying the forces caused to structures by ice. The wall thicknesses are something like 1-4 cm, and they're stiffened pretty good from that point on. I think the approach there has been to keep it simple, low-cost, easy to manufacture and as such sticking with steel and adding some stiffeners to carry the loads and keep deformations elastic was the chosen route.

14. Dec 15, 2004

### pack_rat2

Tell me how thick the wall of a scuba tank is, and its diameter, and I'll tell you how much it deforms, and how to make (if it's possible) a tank that with deform a similar amount when subjected to 30 ksi. I don't think the 0.5 cm value given is correct...it seems way to high. I'll see if I can find some info with a Web search.

15. Dec 15, 2004

### pack_rat2

I found the dimensions, empty weight, and other specs of a 78 ft^3 scuba tank on the Net. They didn't give the wall thickness, but given a spec. grav. of 7.8 for steel, it comes out to around 0.1743 in. That is, in fact, ~0.4 cm, which is more than I expected. The tank is rated at a maximum working pressure of 2640 psi. Doing the calculation for the dimensions given, the "hoop" stress (tangent to the circumference) is 54.9 ksi. If one uses the thin-wall stress formula to calculate the wall thickness required for a tank of similar size, but able to withstand 30 ksi, the thickness obtained is 1.98 in. That is too thick for the thin-wall formula to be applicable, but it's probably not too far off. So you might consider 2 in walls of the same kind of steel as used for the scuba tank, and of similar dimensions, as a guess. I'll check up on a more precise answer...

16. Dec 15, 2004

### pack_rat2

OK, I did the calculation using the THICK-walled formula. This is: stress = [(P*r1^2)/(r2^2-r1^2)]*[1 + (r2^2/r^2)], P is the pressure, r2 is the outer radius, r1 is the inner radius, and r is the average radius. For a cylinder with the same capacity as the one above (ID = 7.25 in, OD = 11.25 in), the hoop-stress is 52.9 ksi...pretty close to the thin-wall result. If you incorporate a 50% safety factor, you need a steel with a yield stress of about 80 ksi...there are plenty of available steels that meet that requirement.

Last edited: Dec 15, 2004
17. Dec 15, 2004

### pack_rat2

Come to think of it, you wouldn't really need much of a safety factor. If this tank were storing a gas under pressure, a large S.F. would be necessary, since if it ruptured, it would do so explosively. But if it contains only ice or water, it would just crack...probably with a loud "clang," but it wouldn't pose a hazard.

18. Dec 15, 2004

### Bystander

Uhhhhh ----

--- wanta rephrase this to "that hazard" ("that" referring to the gas burst hazard)? There're plenty of spalling hazards associated with use of high strength steels in pressure vessels, and plenty of injection jetting hazards around small cracks and failed seals.

19. Dec 15, 2004

### pack_rat2

Picky, picky...;) I wouldn't want to hug it while it was breaking. It would be similar to a tank failing a hydrostatic test by a welding or fire equipment supplier.

20. Dec 26, 2004

### Gara

A normal tin can is able to withstand 13636 tons of pressure??

21. Dec 26, 2004

### PerennialII

Yep, 80 ksi yield strength steel ... take a decent modern steel and all you get is some deformation near 0. Would call this something like "medium" strength.

22. Dec 26, 2004

### Q_Goest

Tanks

Per most state laws, pressurized tanks must meet ASME BPV Code. Since the pressure is well above 3000 psi, Division 2 applies (ASME Section VIII, Division 2). Div 2 rules apply to vessels over 6" ID if I'm not mistaken. The wall thickness for Div 2 vessels is given by:

t = PR / (SE - 0.5 P)
or
t = R ( e^(P/S) - 1)
Where t = vessel thickness
P = pressure
S = Allowable stress (per Code)
E = Joint efficiency (use 1 for seamless vessels)

Note that allowable stress is typically 2/3 yield or 1/3.5 ultimate strength, whichever is less. Note also, the vessel must be manufactured and stamped in accordance with state law by a certified manufacturer.

If the vessel is smaller than 6" ID, these rules typically don't apply. Almost all states have laws which require vessels meet these rules.

If you want to do an experiment in a college setting, using a vessel smaller in diameter than this, I'd suggest using high pressure tubing, first developed by Aminco, now widely manufactured by BuTech, Newport Scientific, Autoclave, and others.
Butech: http://www.butech-valve.com/
Newport Scientific: http://www.newport-scientific.com/
Autoclave: http://www.snap-tite.com/Autoclave_Engineers/index4.html [Broken]

Note that some of these companies also make small vessels which can handle this pressure. Note also, I've not calculated if 30,000 psi is sufficient. That should be done by a PE prior to specifying a vessel if this is an actual experiment.

Last edited by a moderator: May 1, 2017
23. Dec 26, 2004

### brewnog

Are there any ASME standards which specify that a pressure vessel must leak before it bursts (ie such that catastrophic failure is avoided)?

24. Dec 26, 2004

### Q_Goest

You won't find that specifically in the code. But you will find that the materials which are acceptable for use (there is a list specified along with allowable stresses) are ductile materials which will result in the vessel (hopefully) leaking before bursting. Nevertheless, vessels are known to burst on occasion due to fatigue, corrosion, overpressure conditions, etc...

25. Dec 28, 2004

### PerennialII

On most evaluation methods, I believe ASME included, LBB is given as an alternative / complementary method, and officials for example with respect to analyses of nuclear plants accept it as a design measure to certain extent on a case by case basis (or require the LBB criterion to be included, probably differs with different authorities & components). I did a review about this a year or two back and can provide refs if needed.