Is this a daft idea or what?
 

Just wondering if i was to make some ice cubes out of distilled water and was to put 1 or 2 in my resevoir if it would either a) be a totally stupid thing to do or b) make any diiference?

You probably wouldn't get enough cooling from one or two cubes to make a difference. Plus, you'd have to do something with the melt water.

I guess I could drop the amoun in the res?

Given that ice has a lower density than water at the same temperature, so takes up more space for the same mass, if water were removed from the system to allow ice to fit into it, bradmax would actually find it necessary to top up the water once the ice melted.

Given that the latent heat of fusion of water is something like 334kJ/kg, while its heat capacity is only 4.18kJ/(kgK) then the energy required to transfer a kilo of ice to water at 0 Celsius is equal to the energy required to heat that same kilo of water from 0 Celsius to 79.9 Celsius. For a CPU throwing out 100W, this would take about an hour, assuming a purely capacitative system.

If you were to cool water to below air temperature, though, then the radiator would actually start working in reverse - your watercooling loop would pull heat from the air (2nd law of thermodynamics), so your kilo of ice would cool your system for rather less than the 55 minutes - if your radiator were equally matched to CPU output (100W with a 10C delta) then you might be lucky enough to see about 22 minutes in which your CPU runs at close to 10 Celsius.

Of course, if we included heat duty of the pump, entropy of mixing and so forth, the time that the system would be effective is reduced further.

Last time I checked, my freezer could pump only about 25W of heat - it's not a very fast freezer, but will (slowly) chill down to -18 Celsius. Based on that, it would take about 4 hours to chill the kilo of water that my system would melt within 20 minutes...

Incorrect.

Ice floats because it displaces more water by volume than it weighs. However, if I take a liter of water out of a container and freeze it, then put that same liter of water back into the container, the water level will return to its previous position and remain there while the ice melts, because the ice can only displace as much water as it weighs. Since its weight doesn't change when it melts, the volume displaced doesn't change either.

The answer is no :rolling: :rolling:

no dont do it or no its not a daft idea :crazy:

both :D

<pedantic>Ignoring the confusion between mass and weight, you're simply stating that ice is less dense than water. Remember that weight has no meaning without gravity!

Whilst the case you make about the ice floating in the water is indeed correct, that assumes that the water is still not turbulant.</pedantic> Having said that it's not a big deal, as the reservoir will contain more water than is necessary to operate the system. So, neither of you are right. ;)

thought so :D

*sigh* With reference to Perry's Chemical Engineer's Handbook, 7th Ed, R. H. Perry, Don, W. Green, 1997, McGraw Hill; Table 2-1, page 2-27 the specific gravity (i.e. the density relative to water at 0 Celsius) of water at 4 Celsius is 1.00, while that of ice at 0 Celsius is 0.915.

:p

<pedantic> A reservoir large enough for bradmax to consider tipping ice into would NOT be flowing under a turbulent flow regime. Maybe at the reservoir inlet, briefly, but in fact most of the water in even quite a moderately sized vessel would barely be moving.

In any case, care to explain your theory of the relevance of turbulent flow to this problem? </pedantic>

I thought that's what I said? A battleship floats by the same principle, despite the fact that it is made almost entirely from steel. The amount of water that it displaces weighs more than the battleship does. If you took that same amount of steel and just dumped it in a pile in the ocean, it would sink straight to the bottom because in that configuration, its displacement is less than its weight.

And yes, Áedán, I know full well the difference between mass and weight. However, on Earth, they are functionally the same. :p

edit
In any case, I don't see that my original point is invalid. Bradmax was NOT talking about cooling his entire system with a kilo of ice; he was talking about dropping a couple of drink ice cubes from his freezer into his existing cooling system. In order to do that, he would have to remove some existing water from the system (do something with the melt water). Two small freezer type drink ice cubes will not provide enough cooling, IMHO to be worth the effort.

In fact, I doubt that cooling the entire system with ice melt would gain significant performance (MAYBE 5%), simply because the lowest temp achievable for any sustainable period of time would be the temperature at the point of fusion for water, which is 0C at 1 bara. That's not cold enough for the physics of CMOS to really start becoming favorable to overclocking.

Only whilst gravity is a constant. As we well know, the earth's gravity is not consistant across it's surface, therefore weight and mass are not functionally the same. Yeah, ok, it's splitting hairs.

And the difference in gravitational force between the top of Mt. Everest and the bottom of the Marianas Trench is significant only to scientists who measure mass and weight in nanograms. Yeah, it's splitting hairs. :)

No, you're trying to claim that ice is more dense, which is quite simply wrong. You're using an incorrect analogy of a battleship to try to prop up your fundamental mistake.

You are confusing buoyancy with density. You are assuming that ice floats because its shape lends it buoyancy. There you are wrong. Ice floats because it is less dense than water: consider that ice will float regardless of the shape or size in which it is formed. Ice spheres (the shape with the greatest mass/surface area ratio) will float just as well as ice battleships will. Its density lends it buoyancy, not its shape!

Steel, on the other hand, which is clearly more dense than water. Steel spheres will always sink. The only way you can get steel to float is to roll it out into a shape such that its displacement exceeds its mass, as you did correctly state.

I used the example of a kilo of ice to demonstrate that even with a substantial quantity of ice, the effect on the average system would be so small as to be pointless. Therefore using a block or two of ice would be totally ineffective.

In the case of ice, wood, or other materials that are naturally bouyant, the principle still applies. It is bouyant because it NATURALLY has a greater displacement than its mass, and as such can never assume a form that is more dense than water. Steel floats only when it is formed into a shape that CAUSES it to have a greater displacement than its mass, and as such can assume forms that are both more and less dense than water.

In any case, wasn't the original question about how long an ice cube would last in h**l? :rolling:

Hehe yup :drink:

Now if we model hell as a uniform sphere.... ;)

Funny, I could swear there was a response here from Steve, talking about water, ice, and vacuum. But it seems to have disappeared.

There was, but he decided it wasn't worth posting, so he retracted it.