Peltier cooling calculation

[fRiend0]

Hello. I'm currently trying to do a peltier-water all in one cooling with pwm control to avoid condensation. I'm studying mechanical engineering and I want to do this as a project..

However I'm bit stuck right at the beginning of the calculations... Any engineers who have experience with peltiers? I know that they act as heat pumps, transferring heat from cold to hot side when you run some electric power into them... Am I just imagining that it can make the cooling more efficient? 

 

Theoretically speaking, if I overclock the CPU, put a peltier element on top of it and try to pump away the heat, the watercooling will still be my bottleneck and will still transfer the same amount of heat it normally would ... It would just make the peltier element warmer and that's it right? 

[fRiend0]

In that case option to make the water cooling more efficient is to pump out the heat to a bigger surface and make custom block to increase the area from which the water will take away the heat.... Any engineers who could give me some feedback on this?

Thank you

[fjdhdhcyfhf]

cooler master had a peltier product. 200W Air cooling but it is not efficient. No one else copied this idea.

http://www.coolermaster.com/cooling/cpu-air-cooler/v10/

[Organized]

As far I know, peltier elements are way too inefficient to give benefits to any kind of cooling.

[fRiend0]

Well in terms of electric power consumption peltiers are ineficient... in terms of actual cooling when you get to OC temps when your CPU start's to thermal throttle, it can and will bring down the temps... I'm trying to show a proof that with this you will increase the efficiency of cooling... this means that since you are not limited to the actual size of the processor you can make better water flow through the water cooling element... it comes at a price of increased cost of the water block and  higher power bill

[Jumper118]

2 hours ago, fRiend0 said:

Well in terms of electric power consumption peltiers are ineficient... in terms of actual cooling when you get to OC temps when your CPU start's to thermal throttle, it can and will bring down the temps... I'm trying to show a proof that with this you will increase the efficiency of cooling... this means that since you are not limited to the actual size of the processor you can make better water flow through the water cooling element... it comes at a price of increased cost of the water block and  higher power bill

What cpu are you going to cool? 

[sl06bhytmar]

Here you go, I hope you enjoy calculation,

 

You just need to find heat transfer rates for some materials, their thicknesses (or estimates) and surface area size. 

 

I just leave the famous 2600k here and some TDP graphs:

graph 1: shamelessly taken from http://electronics.stackexchange.com/questions/122050/what-limits-cpu-speed

So lets say you go for 5GHz then CPU needs modest 270W cooling power alone.

Lets say you use 200W Peltier element so total you have to transfer to waterblock is whopping 470W. On otherside because you can think peltier as "Heat pump" like you said then on CPU surface is only 70W of power that needs to be still passed trough peltier to waterblock. Basically you just move 400W on peltier hot surface and 70W to CPU surface (might be thinking it wrong atm, haven't done ever peltier heat transfer calculations).

 

Basically you then calculate every heat resistant material separately (those R's in my graph). Then at very end you end up with just

"T_s = CPU's Temperature surface"

T_s = q₁ * R_tot + T_air , where q₁ is your CPU die's heat what needs to be dissipated trough whole system. 

T_s = 70W * R_tot + 21C => yours cpu surface temp.

And what kind heat would be going into your water then? You simply adjust to Ts point to water and calculate it to

 

Twater = (q₁+q₂) * R_tot + T_air,

 

where q₁ is cpu's die's heat what needs to be dissipated to water and q₂ is peltier element's heat that needs to be dissipated to water. You might even get boiling temperatures if you do not have enough rads. NOTE! Rtot is different from previous calculation because materials have moved around in parallel heat transfer plan. 

 

Calculating R_tot = ( 1 / R_{upper area sum} +  1 / R_{lower resistances sum} )^-1

 

To calculate R_die you use formula: Heat resistance = Thickness / (Heat transfer rate * Area) Thus you get
R_die = L_die / ( K_die * A_die ) ,                  { [m] / ( [W] / ( [m][K] ) * [m²] ) } = { [K] / [W] }

example: I assume 2600k's K is 90 W/mK, Die's thickness is 5mm and surface area is 291mm²

 

R_die =  0,0005m / (90W/mK *  0.000291m²) = 0.01909 K / W

 

For convection

R_FC = 1 / ( 100W/m²K *  Rad surface area 90 degrees towards forced convection )

Rad surface area = 0.24m * (1180 fins/m) * 0.025m/fins * 0.12m/12
                                 ^length,    ^fpi converted,  ^rad thickness,  ^rad witdh / how many fin rows

 

Twater = (70W+400W) * R_tot + T_air,

• For CU: 390W / mk {Watts / meterKelvin}
• For FR4: 0.3W / mK
• For Forced Convection: 200W/m²K {Watts / SquaremeterKelvin}
• For Natural Convection: 10W/m²K

 

[Phate.exe]

4 hours ago, fRiend0 said:

Am I just imagining that it can make the cooling more efficient? 

 

Theoretically speaking, if I overclock the CPU, put a peltier element on top of it and try to pump away the heat, the watercooling will still be my bottleneck and will still transfer the same amount of heat it normally would ... It would just make the peltier element warmer and that's it right? 

It's much less efficient, but you can hit sub-ambient temperatures with it.

 

Let's say you have a 125W CPU, and a 75W peltier element.

 

If you just put the heatsink/water block directly on the CPU, you are just moving the heat out of the CPU.  If you just put a heatsink/water block directly on the hot side of a peltier, and don't put anything on the cold side, you're just moving the heat from the cold side (which will frost over) to the hot side.

 

Cooling a CPU with a peltier requires moving the heat from the CPU (125W) , as well as the heat contributed by the peltier itself (75W), so your cooling system must be able to handle the combined 200W cooling load.

[sl06bhytmar]

@fRiend0 check couple posts above, I tried to open it as much as I could without big hassle. (Even more actually)

[fRiend0]

And thank you. You've been of great help :).