Quote:
Originally posted by airspirit
I guess the best way to explain that would be to take a heat source, like a soldering iron, and place it in a bucket of water. The bucket will cool via evaporation, simulating the radiator in this case. When you plug in the iron (increasing the heat load to simulate the increased block efficiency), the water will heat up, and the water will evaporate faster (simulating the greater efficiency of the radiator under greater dTs). Regardless of the fact that the water is evaporating faster, though, the water will be warmer than before. Now, the difference we are talking about isn't as large as turning a soldering iron on, but it is there. There will be an increase in coolant temperature if a waterblock is more efficient.
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I will try and explain with reference to your example.
You are quite correct that when you turn the soldering iron on, there will be an increase in the surrounding water temperature.
However, this is NOT the equilibrium temperature, as at this point, the thermal energy transferred to the air via evaporation of water will be LESS that the thermal energy transferred from the soldering iron into the water.
Because of this, the thermal energy of the water will continue to rise, (so the temperature will rise), and consequently more thermal energy will be transferred to the atmosphere vie evaporation. The temperature will continue to rise until the transfer of thermal energy via evaporation is the same as the transfer of thermal energy from the soldering iron is the same.
So now we have a situation where there is no NET change in thermal energy in the water. The same amount comes in as goes out. This is the equilibrium condition, which is what would happen if you ran a computer crunching for a couple of hours.
The difference in the temperature between the soldering iron and the water is determined by the convective heat transfer coefficient for the soldering iron surrounded by water. This coefficient gives a value of temperature difference required between the edge of the soldering iron and the water which will cause the transfer of 1W of thermal energy. So supposing it is a 70W soldering iron, the required temperature difference between the soldering iron and the water is the heat load divided by the convective heat transfer coefficient .
Note that regardless of the temperature of the water, 70W WILL be dissipated, PROVIDED that the soldering iron manitains this temperature difference above the water.
This is like the efficiency of the waterblock, but there are more stages to consider in the heat transfer, but it is still a coefficient of efficiency.
Improving the efficiency, (ie a lower C/W) will mean that the same 70W is transferred to the water, but a lower temperature difference is required.
AS SUCH, AT THE EQUILIBRIUM CONDITION, THE CPU WILL BE COOLER, AND NOT THE WATER WILL BE HOTTER.
However, as has been said, this does not take into account the secondary heat paths of a fixed efficiency. Essentially, the total heat output of the cpu is conducted away via a number fo routes, each with varying efficiency. By increasing the efficiency of one of them, you will increas the proportion transferred by this route.
However, I still believe that this was not your understanding of why the water will warm up, VERY slightly, as reinforced by your argument above.
So back to your example, once an equilibrium condition has been reached, changing the efficiency of heat transfer into the water WILL NOT change the water temp. The only way to change the water temp is to change the efficiency of heat transfer from the water to the air, as this would change the temperature difference between water and air, OR change the ambient air temperature.
Supposing we have a fixed ambient air temperature, the efficiency of heat transfer from water to air could be IMPROVED by directing a fan at the surface of the water. This will reduce the temperature difference between water and air, BUT the temperature difference between the water and the soldering will not have changed, though due to the water now being at a lower temperature, the soldering iron will also be at a lower temperature.
So I hope you can see from this that, FOR A GIVEN HEAT LOAD, changing the efficiency of a heat transfer stage will ONLY change the temperature delta of the heat producing component above the heat absorbing component. (by heat producing, I mean the component from which the heat is being transferred, so in a water block, the fins are the heat producing component and the water is the heat absorbing component.)
Does this make sense?
8-ball