Open source DIY TEC and Fan controller

[Ultrasonic2]

720watts Qmax is the total of your cooling power from the TECs, this is where your getting confused. For your TECs to be able to cool 720watts of heat you need to cool your TECs with your radiator. So, for your TEC to cool 720watts Qmax, your TEC consumes 600 odd watts of power. What this means is you need a radiator that can cool the 720watts that the TEC transfers from cold side to hot side, plus the heat load created by the TEC from consuming 600watts of power. So what we do is add the Qmax to the power consumed, so you get 1320watts... this figure 1320watts is what your radiator needs to be able to cool down and not have the water inside the radiator exceed a 10* delta from ambient temperature. If your radiator can't handle this heat load you will have a higher water temp in your radiator. The higher the temp the water in your radiator the more of the delta T you gained by adding in TEC's is being chewed up, which results in virtually no difference in temp on the cold side of your TEC chiller under ambient, effectively you have a deltaT of 0*degrees, which means you just consumed 600watts of power for zip, nada, nothing.... good luck to you chap! either you do what i;m doing and learn about TEC's and how they actually work, or just stay away from them.

EDIT: you need to understand some fundamentals of TEC's, they don't "cool" heat as such, they transfer heat from one side of the TEC to the other, you, ie your radiators needs to cool the heat.

actually this is't right either


720watts Qmax is the total of your cooling power from the TECs
Assuming you are applying max amps and volts and keeping the hot side at 26.85c

This is where your getting confused. For your TECs to be able to cool 720watts of heat you need to cool your TECs with your radiator.
Qmax is the max amount of heat the TEC's can move but this is NOT the amount of electricty used to move 720watts from the cold side to the hot side. at full power it's going to use 1512watts



So, for your TEC to cool 720watts Qmax, your TEC consumes 600 odd watts of power. What this means is you need a radiator that can cool the 720watts that the TEC transfers from cold side to hot side, plus the heat load created by the TEC from consuming 600watts of power.
So what we do is add the Qmax to the power consumed, so you get 1320watts... this figure 1320watts is what your radiator needs to be able to cool down



So, for your TEC to cool 720watts of heat, your TEC consumes 1512 odd watts of power. What this means is you need a radiator that can cool the 720watts that the TEC transfers from cold side to hot side, plus the heat load created by the TEC from consuming 1512watts of power. So what we do is add the 720watt (the load your cooling) to the power consumed, so you get 2232watts... this figure is what your radiator needs to be able to cool down


This is where i point out that no ones cpu is generating 720watts of heat so the above is not reality but i just went with it anyway

To me it sounds like EVERYONE should read this

Everything you've wanted to know about TEC's

http://www.tecpeltier.com/viewtopic....7782d7d2b72050

[Liam_G]

actually this is't right either


720watts Qmax is the total of your cooling power from the TECs
Assuming you are applying max amps and volts and keeping the hot side at 26.85c

This is where your getting confused. For your TECs to be able to cool 720watts of heat you need to cool your TECs with your radiator.
Qmax is the max amount of heat the TEC's can move but this is NOT the amount of electricty used to move 720watts from the cold side to the hot side. at full power it's going to use 1512watts



So, for your TEC to cool 720watts Qmax, your TEC consumes 600 odd watts of power. What this means is you need a radiator that can cool the 720watts that the TEC transfers from cold side to hot side, plus the heat load created by the TEC from consuming 600watts of power.
So what we do is add the Qmax to the power consumed, so you get 1320watts... this figure 1320watts is what your radiator needs to be able to cool down



So, for your TEC to cool 720watts of heat, your TEC consumes 1512 odd watts of power. What this means is you need a radiator that can cool the 720watts that the TEC transfers from cold side to hot side, plus the heat load created by the TEC from consuming 1512watts of power. So what we do is add the 720watt (the load your cooling) to the power consumed, so you get 2232watts... this figure is what your radiator needs to be able to cool down


This is where i point out that no ones cpu is generating 720watts of heat so the above is not reality but i just went with it anyway

To me it sounds like EVERYONE should read this

Everything you've wanted to know about TEC's

http://www.tecpeltier.com/viewtopic....7782d7d2b72050

i'm not sure we are talking about the same thing...... he wanted to use 48, yes 48 TEC's at low volts and amps to cool 720watts. at those low amps and volts i approximated a 600watt power usage. so i added the 600watts to the 720 watts to get 1320watts. i'm not talking about the tec i mentioned, he had some other tec entirely.

[Ultrasonic2]

i'm not sure we are talking about the same thing...... he wanted to use 48, yes 48 TEC's at low volts and amps to cool 720watts. at those low amps and volts i approximated a 600watt power usage. so i added the 600watts to the 720 watts to get 1320watts. i'm not talking about the tec i mentioned, he had some other tec entirely.

Oh no. Now we dont understand eachother .


ARRR

[Liam_G]

Oh no. Now we dont understand eachother .


ARRR

and chaos ensues lol.... i'm confident in my situation with the meanwell se600-24 and 6 x 245w tecs...... i shouldn't have tried to rationalize what he was cracking on about

[Ultrasonic2]

and chaos ensues lol.... i'm confident in my situation with the meanwell se600-24 and 6 x 245w tecs...... i shouldn't have tried to rationalize what he was cracking on about

yip it's a sound plan