Cooling efficiency - Noctua 200 on two 240 Radiators

[hannibalxali]

I want to add a noctua 200 to two 240 full copper radiators using custom mountings. Springs from each corner of the fan to the outer corners of the Radiators with suitable screws to fix it in place. My only concern is about the difference between a one 200 mm fan and four 120 mm fans regarding efficiency. Knowing that:

200mm has a 150 CFM, direct contact area is π x r² » π x 10 = 314 Sq cm

120mm has a 107 CFM, direct contact area of four fans is 4 x π x r² = 452 Sq cm

I just multiplied the CFM by the area affected to determine the difference between both methods:

200mm ». 150 CFM x 314 Sq cm = 47100

4 x120mm »107 CFM x 452 Sq cm = 48364

I don't know what are the final numbers but i think that they determine the difference between two methods.. Are my calculations correct? If they were correct then there will be no difference regarding cooling performance. But way less noise with the big fan and probably a better performance in real life, plus the easier control with just one cable and less power consumption of course. Thank you for reading and answering Please see the image provided

[Streetguru]

1 minute ago, hannibalxali said:

 

Why not just use a 200mm radiator?...

[hannibalxali]

Well .. i only found the phobya extreme 200  on the internet

But is not available in Europe and i like the full copper fins in the alphacool one and the reduced thickness as the phobya is 85mm thick! 

 

Also a 240 radiator is a standard size in case of upgrading or even selling and has a bigger area

[hannibalxali]

11 minutes ago, Streetguru said:

Why not just use a 200mm radiator?...

I found one in the UK now .. and i kinda like it .. it's full copper and i can handle the 85mm thickness no problem.. but the thread is still open for answers as i would like to know if my calculations were correct or not.

Thanks for your input

[akio123008]

I'm not completely sure on the maths, but I think having the extra surface area is worth more than having more airflow. For instance 10% more area will give 10% faster heat transfer, whereas 10% faster airflow does not. ( q = h_c A dT   ). (where the airflow (cfm) affects dT)

 

Conclusion: your calculations aren't entirely right but I'm not quite sure by how much, you may still be close.

(not sure how exactly the airflow relates to dT and about the effect of the area not exposed to airflow)

 

You could use the larger radiator if you managed to make some sort of ducting (a small 3d printed/plastic/metal) thingy to expand the fan area to the size of the radiator:

 

[hannibalxali]

55 minutes ago, akio123008 said:

I'm not completely sure on the maths, but ...

Thanks for the info. 

I know my calculations don't describe the whole process of thermal cycle. But i just wanted to study the link of the radiator size with the fan size while taking the CFM in mind.

The calculation shows that the four small fans will beat the big fan by 3% only. 47100/48364 ≈ 97/100 or a similar value if my calculation were not accurate.

I agree with you that the size of the radiator is for sure a critical player in distributing the heat, specially if it was made entirely of copper.

 

I think that 10% more area will transfer more heat but not noticeably faster without a source of pushing force like a fan air.

Thank you for your contribution. I will try to understand the equation you wrote!  

q = h_c A dT 

[akio123008]

1 hour ago, hannibalxali said:

q = h_c A dT 

I have nothing to do so I'll try to provide some more details;

 

q = heat transfer (how good the cooling is, usually measured in watts)

 

h = heat transfer coefficient (some number depending on materials, not really important here as it's constant for both cases)

 

A = the heat transfer area (area of the radiator (this is actually the area of all the fins combined, but as that is linearly related to the area of the rad itself, we can base the calculations on that for simplicity

)

dT = the temperature difference between the air and the radiator or heatsink

 

the airflow caused by the fan causes dT to be greater. The more airfow, the lower the temperature of the air, the bigger the difference in temperature between the hot radiator and cold the air, hence the bigger dT. No fan would mean the hot air doesn't leave the radiator quicly, so dT is much lower, therefore q is very low as well, and the cooling performance sucks.

 

Now we get to the interesting part.

In this equation, the importance of dT and A is the same (they're multiplied together) but dT does not equal airflow. Airflow does indeed affect dT, but I don't think the relation between airflow and dT is linear, meaning that 10% more airflow will result in less than 10% more dT (still more, but not quite 10%)

 

Which means that more area is technically more important than more airflow as 10% more area will (according to the equation) result in 10% better cooling, whereas 10% more airflow will result in say (made up value) 8% larger dT which means only 8% better cooling.

 

1 hour ago, hannibalxali said:

The calculation shows that the four small fans will beat the big fan by 3% only. 47100/48364 ≈ 97/100

Here airflow in CPM and area are multiplied together, meaning the increased importance of area is ignored, meaning your calculation is slightly biased towards the single fan setup. The single fan setup has more airflow, but less area, which is worse than less airflow and more area.

[MaratM]

I do not see any logic in multiplying area with airflow. To be more precise you also have to take fan motor area into consideration. And do not agree with taking max rpm values, at least you have to take airflow values at the similar noise level. Even the greatest (most expensive) nf-a12x25 is loud at 2000rpm but almost silent up to 1100rpm

[hannibalxali]

On 5/9/2020 at 8:33 PM, akio123008 said:

I have nothing to do so I'll try to provide some more details;

q = heat transfer (how good the cooling is, usually measured in watts)...

I appreciate your efforts explaining the equation.

I did a short research on it yesterday, but i can say that your explanation did better.

The actual results will need a real experiment with sensors to major the values in practice. 

I believe that the final watt result after solving the equation is what the heat exchanger system (fan+rad..) is able to manage/time unit. And is related to the watt consumption of the heat source/time unit(gpu/cpu).

When i multiplied airflow by area i used the area that is covered by the fan only. So 200*200mm for the big one and 240*240 for the smaller four fans.

Regards

[For Science!]

3 minutes ago, hannibalxali said:

 

Since ctrl+F didn't yield a result for the term static pressure, I'll just leave it here. Static pressure.

 

While the 200 mm can push a lot of air around in an unobstructed environment, it cannot do it effeciently through an area of resistance (a radiator)

[hannibalxali]

On 5/9/2020 at 9:52 PM, MaratM said:

I do not see any logic in multiplying area with airflow. To be more precise you also have to..

I agree with you.. there are more things to consider and it is not a complete calculation. But it was a starter to ask other members about their opinions and share their knowledge

I'm also very interested in less noise. And i don't find it fair to put the more noisy with the more quiet on the same rack. 

[hannibalxali]

6 minutes ago, For Science! said:

Since ctrl+F didn't yield a result for the term static pressure, I'll just leave...

It also needs a more complicated equation to find out what are the accurate values. but let's talk theoretically

Isn't it because the radiator is taking that airflow and reacting with it?

So i don't think that the additional airflow is going to waste?

[For Science!]

6 minutes ago, hannibalxali said:

It also needs a more complicated equation to find out what are the accurate values. but let's talk theoretically

Isn't it because the radiator is taking that airflow and reacting with it?

So i don't think that the additional airflow is going to waste?

Radiators work efficiently because the air travels through it, carrying heat away from copper fins and exiting on the other side (thereby not re-introducing it into the radiator). If you have weak static pressure, the air just kind of bounces around and also has the chance to leak back out via the peripheries of the fans. This allows for the heated air to be re-circulated into the radiator (it won't heat up the radiator per se, but the warmer air will be less efficient at removing heat).

 

In your particular case, the if you have massive gaps around the 200 mm fan, you have an additional path of lower resistance coupled with low static pressure, so the air will likely lick the surface of the radiator and then proceed outwards laterally instead of through the radiator. This means that the majority of the copper fin surface area the radiator spent heating up never sees good airflow.

 

I think it's a highly complex situation and probably requires some kind of fluid dynamic simulation to grasp properly,

[hannibalxali]

There are many best answers on this post! 

It would be better if i could choose multiple ones