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[Solved] Peltier/WC Hybrid question (thermal dynamics)

Forum Overclocking : Heatsinks & Air Cooling [Solved] Peltier/WC Hybrid question (thermal dynamics)

btw this is gonna be a long winded explanation

Ok, so I've been trying to do some research and cant seem to find quite what im looking for as it relates to my questions. Basically I need some help figuring heat load and heat dissipation/absorption.

I'll start by trying to explain what im wanting to do. Im trying to plan out a sort of hybrid water cooled peltier system for my pc. Its crudely similar to the Coolit Boreas. http://gear.ign.com/articles/832/832080p1.html But in my idea I dont plan on using multiple blocks for the coolant to pass thru. Imagine if you take a square reservoir, cut the back off it and slap the cold side peltier cooled heatsink, then use the cooling effect of the cold heat sink to absorb the heat from the fluid. Basically it would be like an active radiator vs a std radiator with fans.

Now, what I need help with figuring out (understanding) is how to figure up the total heat generated by my cpu/gpu/etc that is being stored in the coolant AND how to calculate surface area at "X" temperature to effectively absorb the heat in the coolant.

My theory is that if I can figure up how much heat is being generated I can figure how much I need to cool it. Since heat sinks are usually rated in dissipation I thought I could get a somewhat relative figure to calculate surface area to dissipation ratio. Then use that ratio to somewhat figure how much surface area it takes to pull 1 degree from the coolant...

rough numbers (and probably wrong ) :for example, If I cool the peltier heat sink to 20c and the sink has a 200"3 surface area and I get a 5c drop is coolant, would it be correct that to double the surface area of the same I would get a 10c drop? Is my logic making sense or am i just rambling?

IF can calculated I think I get my devices to the chilly temperatures of the peltier without all the hassles of condensation. Does this make sense to anyone?

Ahhhhhh, now someone pass me and oxygen tank please...

http://img716.imageshack.us/i/peltierdesign.jpg/]

Message quoted 1 times
Message edited by wolfman123bugs on 03-06-2010 at 05:03:15 AM

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Best answer

wolfman123bugs wrote :

If I understand you correctly, you are basically doing this: CPU --> Water --> Peltier, correct? If it is, you are basically creating a water chiller. That's not a bad idea, but you need to figure out how you are going to transfer the heat from the Peltier into the air, else your Peltier will fry in short order. I bet you will have to water cool the hot side of the Peltier. Personally, I would just say if you want to have a water chiller, get a pre-made chiller with a compressor as it will be less hassle than a water loop - Peltier - water loop setup.

Quote :

Now, what I need help with figuring out (understanding) is how to figure up the total heat generated by my cpu/gpu/etc that is being stored in the coolant AND how to calculate surface area at "X" temperature to effectively absorb the heat in the coolant.

The total heat generated is very close to the amount of power the components draw. You will need to find and sum the power dissipation of the CPU, GPU and everything else in the water loop and then add to that the power consumption of the water pump (as they do warm the water slightly.) If you want to know how much heat you need to pull off the Peltier, take the previous number and add to it the power consumption of the Peltier.

Quote :

My theory is that if I can figure up how much heat is being generated I can figure how much I need to cool it. Since heat sinks are usually rated in dissipation I thought I could get a somewhat relative figure to calculate surface area to dissipation ratio. Then use that ratio to somewhat figure how much surface area it takes to pull 1 degree from the coolant...

rough numbers (and probably wrong ) :for example, If I cool the peltier heat sink to 20c and the sink has a 200"3 surface area and I get a 5c drop is coolant, would it be correct that to double the surface area of the same I would get a 10c drop? Is my logic making sense or am i just rambling?

IF can calculated I think I get my devices to the chilly temperatures of the peltier without all the hassles of condensation. Does this make sense to anyone?

Ahhhhhh, now someone pass me and oxygen tank please...

http://img716.imageshack.us/i/peltierdesign.jpg/]

There are equations that you can use to find the surface areas and temperatures in a forced convection (which is anything with moving air or liquid) system. Here are some of those equations. I can help you run them if you can provide some rough numbers as to the parts you use and your room temp.

Message edited by MU_Engineer on 03-09-2010 at 02:02:20 AM

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