Hey 8-Ball and Alchemy, when you recover from your hangovers take another stab. If nothing else, you guys are helping me get a better handle on this stuff.
Although Graystar is phenomenally obstinate about holding onto his ignorance and misunderstanding, there is some evidence that he may actually be able to acknowledge (at least to himself) that he is wrong, from the fact that he hasn't posted in
this thread in a few days. (Sorry, my thought processes seem to consist primarily of run on sentences, and parenthetical expressions)
Quote:
Originally posted by Graystar
What I was saying is that you can't keep the core hot for free.
The CPU is at a temperature above ambient and remains there so as long as the computer is on. Some thermal energy is used in maintaining that elevated temperature.
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Once again you are displaying fundamental misunderstanding.
Ever put coffee in a thermos and drink it a few hours later? Still hot? Hmmm. Where's the energy input that maintained the temperature of the coffee?
Unless there is a path for thermal energy to leave an object, it will maintain its temperature. If thermal energy is continuously added to an object with no 'heatsink' the object will continually increase in temperature.
Quote:
Originally posted by Graystar
So when a processor dissipates 70W of power in one hour, I'm saying that 238 BTUs are removed by water cooling, and 1 BTU is used keeping the core hot. (values are demonstrative only)
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You seem to have 'power' and 'energy' confused.
Power is a
rate of energy usage/dissipation/flow/whatever. So it would make sense to say, "When a processor use 239 BTU's (70 Watt-hours) of energy at a constant rate for an hour, the
rate of energy usage is 70 Watts."
Yes there is greater thermal energy in a processor that has been running for some time, than in one that has not. That elevated energy level is 'achieved' in the process of reaching thermal equilibrium. Once equilibrium is attained though, no further energy is "used" in maintaining that elevated energy level. (and corresponding temperature)
At equilibrium the flowrate of electrical energy that is dissipated in the CPU must be equal to the flowrate of heat energy leaving the CPU, otherwise the energy in the CPU would continually be climbing, and the temperature would be climbing with it.
To go back to your example:
If in one hour 239 BTU's of electrical energy were dissipated in the processor and only 238 BTU's of thermal energy left the processor then it is not at equilibrium yet. Are you suggesting that after the next hour there are 2 BTU's left, and then 3, 4...?
Graystar, I know that you understand that the coolant flowrate is the same through all components of a water cooling loop. Is it such a stretch to understand, that the energy flowrate is the same through all the components of a thermal system at equilibrium?
Think about it.