[8-Ball]

I think I get the idea.

In your case, you have a liquid coming into a heat exchanger, that you want to heat up. In this case, yes the liquid needs to remain in the heat exchanger for a longer period of time, so that it will absorb more thermal energy. In effect, you are looking for a maximum temperature differnce between the liquid before and the liquid after the heat exchanger.

However, in the case of water cooling, we are not concerned with the temperature differnce of the water. An ideal coolant would have an infinte specific heat capacity, and the temperature would not change at any point in the loop. We are concerned with the temperature of coolant required to dissipate the heat.

So in your case, you need to get the liquid to as high a temperature as possible, but this has to be in a single pass through the heat exchanger. If you speed up the flow, you still want the same thermal energy to be dissipated into the same volume of water. Obviously, as you have stated, this is not possible.

But for water cooling, if we speed up the flow, then we don't need to dissipate as much per unit volume of coolant.


So, in your case, if you want to double the flow rate, and have the liquid coming out at the same temperature, then you need to double the thermal energy dissipated by the heat exchanger, however, the heat dissipation in watercooling is constant, so the flow rate doesn't matter in the same way.

I hope this makes sense. It's quite hard to explain.

8-ball