Watercooling performance  and amount of water or liquid in the loop Imagine a watercooling system. It consist of a reservoir, pump, tubes and a radiator. We all know how it works.
My question is, only inreasing the reservoir's volum, say from 1 liter to 2, will actually increase the performance?
I do remember some thermodynamic laws from school, Delta T of water to ambient after a given time will be same? Or becaue there is more water in the system it will perform better?
 A: The volume of the reservoir does not affect the performance directly.
The performance (heat taken from the device being cooled down per second) is determined mainly by the temperature of the water. This temperature must not increase and this should be ensured with minimal waste of energy.
The water in the cooling system goes the following loop:


*

*The water of temperature $T_\text{min}$ comes to the radiator.

*In the radiator the water takes energy from the device being cooled and its temperature rises to $T_\text{max}$. Let's denote the integral thermal conductivity of the radiator as $\varkappa_\text{rad}$.

*The water of temperature $T_\text{max}$ leaves the radiator and contacts with 
the environment (usually with the atmosphere) of temperature $T_\text{env}$. The integral thermal conductivity of this part of the system is $\varkappa_\text{env}$. The temperature of the water changes to $T_\text{res}$ during this process.

*The water of temperature $T_\text{res}$ comes to the reservoir. If the reservoir has some additional cooling system (characterized by $\varkappa_\text{res}$) then the temperature of the water decreases to $T_\text{min}$, else $T_\text{min} = T_\text{res}$.


The thermal conductivity values here are integral values that take into account the fact that temperature of water is not constant inside each radiator.
The thermal balance equation is
$$
Q_\text{rad} = Q_\text{env} + Q_\text{res}
$$
or
$$
 (T_\text{max} - T_\text{min}) =
 (T_\text{max} - T_\text{env}) +
 (T_\text{res} - T_\text{min}).
$$
The efficiency of main radiator $\varkappa_\text{rad}$ should always have maximal value. For $\varkappa_\text{env}$ and $\varkappa_\text{res}$ we have to consider different cases.
Water is cooled by the environment
$$
T_\text{min} = T_\text{res} = T_\text{env} < T_\text{max} 
$$
The temperature of the environment is low enough for effective cooling.
We don't need waste energy for additional cooling and can make $Q_\text{res} = 0$ and $\varkappa_\text{res} = 0$.
The value of $\varkappa_\text{env}$ should be maximized as well as $\varkappa_\text{rad}$.
The volume is not important here. We need maximal area of the radiator's surface. If the volume is constant, increase of surface leads to high tension decreasing the flow through the main radiator and overheating. Hence the volume should be large enough but adding a barrel in the middle of tube will not help much.
One can make the external radiator as large as he wants or even use a nearby lake.
Water is cooled inside the reservoir
$$
T_\text{min} < T_\text{res} = T_\text{max} < T_\text{env}
$$
The environment is too hot for the device and we need to exclude it from the system: $Q_\text{env} = 0$ and $\varkappa_\text{env} = 0$.
The value of $\varkappa_\text{res}$ should be maximized as well as $\varkappa_\text{rad}$. But if the reservoir is big the heat exchange with the surrounding is also high.
The reservoir should be big enough to contain proper cooling system but not larger.
Water is cooled by everything
$$
T_\text{min} < T_\text{res} = T_\text{env} < T_\text{max}
$$
The environment is colder than the device but is not cold enough to provide proper cooling.
The water will be cooled in two stages:


*

*cool it to the environment's temperature to get better starting conditions for the next stage of cooling and save energy.

*cool it water inside the reservoir


Here both $\varkappa_\text{env}$ and $\varkappa_\text{res}$ should be maximized and two reservoirs are required.
The first one can be as large as we want (for the exchange with the environment) but larger than minimal size needed for heat exchange.
The second one should be as small as possible just to provide enough cooling.
Conclusion
The volume of the reservoir does not affect the performance directly. The only requirement is: there should be enough water for the flow we want.
