This is wrong. May I ask where you were taught thermodynamics and fluid mechanics?
There's a few places where
@cmaier was maybe a little loose with terminology, but you're gonna have to be a bit more specific if you want to claim he's all wrong. Because he's not.
The relevant high school physics equation is Q=c*m*dT. Q is heat energy (often measured in Joules), c is the "specific heat capacity" of the piece of matter you're analyzing (this is a number which varies based on material properties, but is constant for any given material), m is the mass of the material, and dT is delta-T, the temperature rise that material experiences if Q joules of thermal energy are transferred into it. (If Q is a negative number, that's joules moving out AKA cooling, and dT becomes a negative number too.)
Now let's talk about power. Power is a measure of energy over time, and the SI unit is the watt. One watt is one joule per second.
If a chip is consuming 100W of electrical power (100 J/s) and converting it to heat as a byproduct of doing some computation with it, 100 joules of thermal energy are being added to the die every second. If you assume there is no outflow of thermal energy from the die, for every 1 second it spends in this condition it's getting 100J of thermal energy added to its mass and the temperature of that mass will go up by the rearranged equation dT = Q / (c * m) = 100 J / (c * m) where c is 0.7 J/(gram * degC), the specific heat of silicon. (I'm ignoring the mass and properties of all the non-silicon materials in the active layer of the die, here, because their mass is really low compared to the bulk silicon of the die.) If we assume the die is 1 gram, that'd be 100/(0.7 * 1) = 142.9 degrees C temperature rise every second.
If that 100 J/s influx of thermal energy is balanced by a 100 J/s outflux - by passive and active cooling providing a net -100W thermal power to the die - the die's temperature neither rises nor falls.
If there's a 100W influx and a 101W outflux, the die's temperature will fall.
That's what
@cmaier is talking about with the lake analogy. The difference between energy influx and outflux determines how rapidly and in what direction temperature changes, just as a lake fills or drains at a rate proportional to the difference between the amount of flow into and out of the lake.
Now let's think about an artificial lake created by damming up a valley. You don't ever want the lake to top the dam, because that usually results in a dam failure. So whenever the lake starts getting too full, the people running the dam (or an automatic system) open a valve to let water flow out at a higher rate.
That's exactly what Apple's thermal control loops are doing. They don't bother goosing the outflow (ramping up the fans) until temperatures rise somewhere close to safe operating limits (I'm using safe to imply computational correctness - see
@cmaier's earlier posts). Then, to keep "water" from "overtopping the dam", they raise fan speeds until thermal energy inflow and outflow are balanced and the chip's temperature stops rising.
Is it possible to keep die junction temperature lower by running the fans faster earlier so that equilibrium is reached by the time the chip hits 60C, or whatever? Sure. Do you need to? Nah.
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