What is the maximum pressure attainable with expansion of water to ice in a closed container?

Question

If you have a steel container with 2" thick walls, you fill it with water and weld a 2" plate on top you will have a strong container completely full of water.

If you then freeze this container how much pressure would be generated inside? Will the water still turn to ice? I know engines have cracked due to a lack of anti-freeze in winter and mountains are cracked via freeze-thaw.

Any ideas appreciated.

Answer 1

I looked at those graphs, and, quite frankly, was unable to figure out the answer.
http://www.lsbu.ac.uk/water/phase.html
The question implies that volume is constant, but the graphs do not show any volumes. They are P-T diagrams, not P-V diagrams.


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Water enclosed in rigid shell will follow red path
A -> B -> C-> D
as you decrease temperature.
http://i23.tinypic.com/f9qudd.jpg

From point A(0C) to point B(-22C) water will remain mostly liquid, as pressure goes up from almost zero to 207MPa ~ 2,000 atm.

Small part of water, however, will crystallize into ordinary ice. The part is small, because water is neither compressible, nor has much of thermal contraction of volume.

At point B all water will freeze, part as ordinary ice Ih, and part as ice III.

Further decrease of temperature will cause slightly more increase in pressure, as you go along line B->C, and finally reach maximum pressure 213 MPa (~2,100 atm) at point C(-34.7C).

Answer is: maximum pressure = 213 MPa @ -34.7C

At point C ice III will re-crystallize into ice II, and further cooling will cause slight decrease in pressure.

Answer 2

(corrected)
I'll simplify things and assume the container is maintained at constant volume (Your passive steel container won't qualify). The best graph to answer this question on Jack's link at http://www.lsbu.ac.uk/water/phase.html is the 5'th from the top. The liquid starts out at 1 g/cc. If we maintain that density and cool to -10 C (the freezing point drops with pressure), you get roughly 200 MPa (not the 2 MPa I said early; Jack *did* get the right pressure; I misread the log scale initially) . The vertical line at the transition crossing 1 g/cc means it will form a mixture of about 70% Ice 1h and 30% Ice III, corresponding to different crystal structures. In a real passive container, the pressure will be less.

Answer 3

You can use the ideal gas law if you want a loose approximation. If you want something more accurate, use Van der Waals, Redlich-Kwong, or the Virial equation of state. From the standpoint of the ideal gas law, first determine the volume V of the container and the number of moles n of H2O.
Then as P=nRT/V,
Pf - Pi = (nR/V)*(Tf-Ti),
so the pressure difference is directly proportional to the temperature change, with a scalar multiplier of value nR/V.

Answer 4

http://www.lsbu.ac.uk/water/phase.html

Look above 10^8 Pascals at 273.15K (0C). Also, the second graph, just below it should be very helpful.

Notice also, how ice II, III, V, and VI have a higher density than water. Which means they have a lower volume as a solid than they do as a liquid, unlike ice I.
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