New to watercooling, looking for feedback

[tachi1247]

Yes. In fact, even the fan speed doesn't matter. That 1 or 2 C is the difference in temperature between the inlet and outlet of the radiator. That doesn't say anything about the amount of power (heat/time) that is being dissipated....

That last part is definitely not true. If you know the inlet temperature, outlet temperature and the flow rate (volume) of the coolant you can calculate exactly how much heat (power) is being dissipated by the radiator.

...or the difference in temperature between the coolant and the air entering the rad.

You are correct that we can't determine the difference between coolant temp and air temp, but that isn't really relevant to anything.

Turn the fans up and the system will run cooler, but you'll still have the same differential between inlet and outlet, because that is determined only by the flow rate and how fast heat is dumped into the system. If you increase the speed of the pump, though, the the difference between the inlet and outlet will be smaller.

There are 2 things that effect the temperature differential seen on either side of the radiator (assuming you keep the ambient air temperature a constant). One is the flow rate and the other is the efficiency of the radiator (which is a heat exchanger). A faster flow rate will result in higher outlet temps than a lower flow rate because that specific volume of coolant has less time to be cooled by the radiator. Fan speed works differently in that it affects the temperature differential that the radiator can generate. Increased airflow over the radiator will increase it's efficiency, but there is a point of diminishing returns on increasing fan speed. Each radiator is going to have a sweet spot where it will be most efficient. A low fpi radiator would have a much higher outlet temp if you ran the fans at 200 RPM compared to 1000 RPM. 1000 is probably near that sweet spot so increasing another 800 RPM to 1800 probably wouldn't change outlet temperatures much if at all.

We could dissipate any amount of power with a small temperature differential between inlet and outlet if the flow rate was high enough. Consider the following examples.

With 100 watt load and a 1 GPM flow rate, we'd have a 0.38 C differential across the radiator. Using the math iRemainStanding presented, that is how much warmer 100 watts would make one gallon in in one minute.

200 W at the same flow rate would raise that to 0.76 C.

Increase the flow rate to 2 GPM, the we're back down to .38 C.

If we needed to cool 800 W and our flow rate was only 1 GPM, we'd have 3 C between the inlet and outlet of the radiator.

But the pumps in an insanely powerful system like that would probably push at least 1.5 GPM (because whoever built it obviously believes that if some is good, more is better), so the difference across the radiator(s) would be only 2 C.

See how it works? Measuring 1 C across the radiator doesn't tell us anything about how fast heat is coming out it.

If we took the radiator out of any of those systems, most of the heat added to the loop would remain in the coolant. Let's assume the reservoir and tubing hold in our waternotcooling system hold one gallon of water. 100W would raise that .38 C per minute. Ignoring losses from the tubing and res, the coolant would reach be 38 C over ambient in 100 minutes. Obviously we need a radiator.

So what happens if we use a radiator half as large? The differential across the radiator inlet and outlet is still only .38 C for 100 W and 1 GPM, because we still have the same amount of power and the same flow rate. But the difference between the coolant temp and the ambient air temperature will double. Double the radiator, and the difference between coolant and air is cut in half. This is why radiators like the Phobya Xtreme SUPERNOVA 1260 and the Watercool MoMO-RA3 9x140 exist.

Is it starting to make sense?

This part all makes sense. I was just saying that if you really only need a 1 degree temp drop in the coolant to maintain equilibrium at an acceptable temperature, then you could probably achieve that by increasing the total volume of water in the system. The examples above ignore heat loss through the tubing and of the water cooling via ambient air in the reservoir and obviously in the real world that heat loss occurrs. If you had enough water (might be several gallons or more) in the system you could theoretically achieve that same amount of heat loss without the radiator.

After all of that, I do get it. While I still believe there is a theoretical "best" way to setup the loop and that putting the radiator before the GPU would lower the temp the GPU sees, it seems the practical value of doing so is insignificant. Basically it comes down to the flow rates, heat being producted, etc of the typical system result in a real life scenario where the order of components doesn't matter. Thanks for talking it through with me though....

 

[Otter]

Yes. In fact, even the fan speed doesn't matter. That 1 or 2 C is the difference in temperature between the inlet and outlet of the radiator. That doesn't say anything about the amount of power (heat/time) that is being dissipated..

That last part is definitely not true. If you know the inlet temperature, outlet temperature and the flow rate (volume) of the coolant you can calculate exactly how much heat (power) is being dissipated by the radiator.

And if you don't, the difference between the inlet and outlet temperature alone doesn't tell you much. Hence, you can't determine whether or not the radiator is needed based only on that differential.

You are correct that we can't determine the difference between coolant temp and air temp, but that isn't really relevant to anything.

It's relevant to the idea we don't need a radiator because it's only cooling the water 1 C.

Increased airflow over the radiator will increase it's efficiency, but there is a point of diminishing returns on increasing fan speed.

Of course. There is a point of diminishing turns to everything. Taking it to an absurd extreme, a strong enough flow of air could actually heat the radiator through friction.

This part all makes sense. I was just saying that if you really only need a 1 degree temp drop in the coolant to maintain equilibrium at an acceptable temperature, then you could probably achieve that by increasing the total volume of water in the system.

Well, yes you could. But how big is your reservoir going to be? To determine that, you need to look at the power being dissipated, not the temperature drop across the radiator.

The examples above ignore heat loss through the tubing and of the water cooling via ambient air in the reservoir and obviously in the real world that heat loss occurrs. If you had enough water (might be several gallons or more) in the system you could theoretically achieve that same amount of heat loss without the radiator.

And at that point, the reservoir becomes a radiator. Back in the days of when a high-end CPU dissipated 25 watts, and most watercoolers made their own blocks, there were some systems like this. Some used five gallon reservoirs. I recall one very intersting rig that ran the coolant through hollow copper side panels. You'd need a very large res to cool a typical watercooled gaming rig today, but the Zalman Reserator is still kicking around. I refer to the big, finned reservoir, not the "MAX" Reserator which has an automotive style radiator and fans.

After all of that, I do get it. While I still believe there is a theoretical "best" way to setup the loop and that putting the radiator before the GPU would lower the temp the GPU sees, it seems the practical value of doing so is insignificant. Basically it comes down to the flow rates, heat being producted, etc of the typical system result in a real life scenario where the order of components doesn't matter.

Exactly. But it's good to understand why it doesn't matter.

Thanks for talking it through with me though....

Glad to help.

 

[iRemainStanding]

I took a trip to Microcenter to exchange my GPU for a different one that would support a waterblock and they had an openbox refurbished R9 295x2 for $510. For an extra ~$170, I figured why not. Also got a corsair RM 1000w while I was there. New question now: will the rads I have chosen be enough to cool all this, or will I need to get thicker ones?

 

[GTXJackBauer]

I took a trip to Microcenter to exchange my GPU for a different one that would support a waterblock and they had an openbox refurbished R9 295x2 for $510. For an extra ~$170, I figured why not. Also got a corsair RM 1000w while I was there. New question now: will the rads I have chosen be enough to cool all this, or will I need to get thicker ones?

Yup, you should be good.