[Sirpent]

Originally posted by schoolie:
Assuming that the dissipation of heat from the water to the radiator is proportional to the difference in temp between the water and the radiator ( and the air flowing over the fins). If one accepts this, then the heat dissipation would increase with increasing water temps in the radiator, and with lower air temps flowing over the radiator fins.

That's a reasonable assumption (of course, we leave out turbulence here). The radiator itself is not important. We can assume that the heat transfer is proportional to
(Tair -Twater), where Tair = (Tair-in +Tair-out)/2, Twater = (Twater-in +Twater-out)/2.

Consider the steady state of a simplified system, where the coolant temp is only a function of position in the cooling loop, and not time dependent.

That's fine.

A lower coolant flow should produce a larger difference in inlet and outlet coolant temps in the radiator, and a lower total heat dissipated by the radiator.

Not necessarily. With lower flow Twater-in will go up improving heat transfer. It's not that your answer was wrong, just the argument was not convincing.

I'll finish my thoughts after I get back, but I'd be curious what other people think.

My 2 kopecks.

Well, the problem is, we have much fewer equations then variables in our system. In the real life, there are additional relations between those variables but they are much more complex. It seems hard to get reliable answers from this model. It should be easier to analyse the system mathematically in the "limit case", where the flow is very high and (Twater-in -Twater-out) is very low.

Originally posted by bigben2k:
So how does that explain the sweet spot of rads, as per the link above?

Frankly, that sweet spot phenomenon looks weird. If the measurements are correct, the only possible explanation is that the "volume" of turbulence behaves in a non-monotonic way: water turbulence in the radiator has "resonances" at certain values of the flow.

You see, a "sweet spot" in a real life system is conceivable because the waterblock is also involved in some obscure way. But Bill kept (Twater-in -Tair-in) constant, and this is an entirely different story.