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When water freezes in a pipe it can crack the pipe open. I assume this takes quite a lot of energy as when I try to crack a pipe it can be hard work!

I think water freezing is a result of energy (heat) being lost from the water and out of the pipe into the freezing environment around it.

So what energy is cracking the pipe and how? When warm and not frozen there is more energy in the pipe than when frozen?

My secondary question might be - is this a particular phenomenon of water or would other matter crack open a pipe when it freezes solid from liquid?

Andrew

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Since your two questions are different, they can go in different threads. See meta.physics.stackexchange.com/questions/13/… Good question, though. –  Mark Eichenlaub Jan 17 '11 at 8:05

3 Answers 3

up vote 8 down vote accepted

First of all, when you say that trying to crack a pipe is hard work, what you probably mean (in physics terms) is that it takes a large force. But that doesn't necessarily mean that it requires a lot of energy. The energy used in a physical process like that is equal to the force times the distance over which the force is applied, and you don't have to push very far in order to crack a pipe. In fact, the pipe barely moves at all before it cracks, so even though the force required is quite large, it only acts over a tiny distance, and therefore it barely takes any actual energy. What little energy is required can come from the water itself.

To explain the "how" you have to consider molecular interactions. (Well you don't have to, but I'm going to.) The energy of each pair of water molecules varies with the distance between them, in a manner shown (approximately) by the following graph (from Wikipedia).

Lennard-Jones potential

You'll notice that there is a certain distance at which the energy is a minimum. This distance represents the "natural" equilibrium distance between molecules when there is no pressure. However, when the water is under pressure, the molecules get pushed together (because pressure is roughly akin to force), so their actual distance will be a little closer than the minimum of the graph.

Water has the unique property that its "natural" density at a constant pressure reaches a maximum at a certain temperature, around 4 degrees Celsius, and that its frozen form (ice) is less dense than its liquid form. In other words, the equilibrium intermolecular distance (the minimum of the graph) is smallest at 4 degrees Celsius. If the water temperature is going to drop below 4 degrees Celsius, the minimum shifts a little to the right, which means one of two things has to happen either the water expands (if the intermolecular separation stays with the minimum of the graph), or its pressure rises (if the intermolecular separation creeps up the slope of the graph).

Now think about the situation in a pipe. As long as the pipe stays intact, the water can't really expand at all. So the only option is for the pressure to increase. As the pressure increases, the force on the pipe increases as well, and you'll notice that because the slope of the curve is very steep, the force increases very rapidly. At some point, the force becomes large enough to overwhelm the bonds that hold the atoms/molecules in the pipe together, and at that point, the pipe cracks. Notice that in this theoretical model there's no need for any part of the pipe to have moved, which means the pipe could crack without any energy being used. (In practice, there are other things going on that do make it take a tiny bit of energy.)

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Good answer. I'd be interested also in precise dynamics though. In reality there are usually some inhomogeneities in the pipe and this is what the ice exploits, tearing the pipe in the weakest spot. But if we assume a pipe with both rotational and translational symmetry then I have no idea what would happen. Do you think this might constitute a good question? –  Marek Jan 17 '11 at 9:30
    
@Marek: sure, I guess so, although it seems to me that there would be some sort of spontaneous symmetry breaking (no pun intended) causing a weak point to appear. If you can think of a reason why there might be more to it, that would definitely be a good question. –  David Z Jan 17 '11 at 10:08
    
@David If You started with "Water has the unique..." and omitted everything before (including the misleading diagram) the answere would have been perfect. Water ice has a diamond-like structure due to hydrogen bonds, which is reason for its lower density as compared to liquid water. –  Georg Jan 17 '11 at 11:51
    
@Marek,Hmmm in case of that mechanical ideal pipe one would have to resort to some thermal fluctuation or sound wave from the whisteling experimenter or what ever You want to cause the rupture somewhere. Georg –  Georg Jan 17 '11 at 14:55
    
@Georg (2 comments up): starting with the third paragraph would have meant omitting the actual answer part of my answer, as well as a fair amount of prerequisite knowledge. And even though the crystal structure of ice does arise because of hydrogen bonding, qualitatively the potential still looks roughly the same (in that it has a minimum) and the graph still applies. There just happens to be some odd behavior in the position of the minimum as the water freezes. –  David Z Jan 17 '11 at 19:38

I do not claim to understand all the numbers (I'm just a public defender, not a real lawyer) but I think this particular problem is based on more than a simple calculation of the volume/density change in water as it freezes. Pipes are not usually closed systems; for instance, any collected water in drain pipes is free to expand along the length of the pipe as it freezes and so the chance of a drain pipe bursting due to water freezing is minimal. Similarly, the water in a water supply pipe is normally only under the pressure of the incoming water supply (from a well or a much larger underground city water supply pipe). This amount of pressure is well within the pipe's ability to resist—in fact the pipe is designed to do so. The only problem arises when the water expands when it freezes.

But it is very important to know where the water freezes first. If the water freezes in an outdoor faucet for a garden hose FIRST, it probably won't burst the faucet or the pipe. During the phase change the water is free to flow a little bit and so the faucet does not even have to deal with the full pressure of the water that expanded within it (approximately 10% by volume if I'm reading the above posts correctly); as in the drain pipe example, the water is free to expand along the length of the pipe—all it has to do is force a little of the water supply to back up by overcoming the relatively low pressure of the water supply. This pressure relief phenomenon is likely to occur whenever water freezes first at one of the many ends of a plumbing system. Therefore, most burst pipes do not occur simply from ice itself.

You see, the real problem occurs when the water first freezes solid somewhere between the water supply pressure and the end of the pipe, creating a blockage. When that happens, you end up with water trapped between the frozen water blockage and the end of the pipe. That water has nowhere to go when more water starts freezing between the frozen blockage and the end of the pipe—the water pressure just rises and rises until it bursts the weakest part of the pipe and escapes. So in the end it's not really the ice that bursts the pipe, it's the water that hasn't frozen yet. I can tell you that my plumber friend agrees with this explanation and it helps to explain why larger pipes are less likely to burst (because it's much harder to have all of the water in a section of pipe freeze solid at once and create a blockage).

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Welcome, Sam and thank you for this post! However, please note that posts here usually don’t contain greetings or signatures (you have a little signature below each post of yours). ☺ –  Claudius Nov 15 '12 at 20:00

A very simple answer is that by freezing water increases its volume or decreases its density from 1000 kg/m^3 in the liquid phase down to 920 kg/m^3 in the solid one. Details you may find here . One can easily calculate out of here the increase of volume that would experience the mass of a free water (equal to that inside the tube). Since for the same mass, m, of water one finds m=\rho _wV_w=\rho _iV_i. Here the subscript w indicates water, while i points out ice; V is the volume and \rho is the density. Now from this it is easy to find \frac{V_i-V_w}{V_i}=\frac{\rho _w-\rho _i}{\rho _w}=0.08 This is a huge strain. The bulk modulus of water is about B\approx 2 \text{GPa} (you will find it in the same Wikipedia article). Then the pressure in the closed pipe is P=B\unicode{f39e}\times \text{$\Delta $V}/V. This yields P=1.6\times 10^8 \text{Pa}. This pressure breaks the pipe. Note that it is only 2-3 orders of magnitude smaller than the theoretical strength of metal which is \sim 10^{11} \text{Pa} , but as you know, solids yield fracture way before their theoretical strength is achieved.

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