Help me understand TECs a little more please.

Ok I been reading the guide on TECs.
I understand some of the math.

I been looking at a 400W Tec. It runs on a max of 15.4V.
I will be running it on 12.00V.
So that would give me 311.69 Watts of cooling.






Now a Q6600 Over clocked to 4.0Ghz using 1.55V should pull around 243W.

So Converting Volts to Amps at fixed wattage.
watts/volts = amps on the TEC
bout 26Amps on the CPU and 25Amps on the Tec.


So what do I do with the math from here to make sure Im in the clear?
what other math do I need to do?
Know what kind of cooling power I will get and how much power it will take to run the TEC on 12V.

Thanks

bump for some help with math plz.

so the max temp difference is i think only -15C does that sound right?

is my math right?

bump.

Um... Did you read this;
http://www.ocforums.com/showthread.php?t=309722
Sorry I can't be of more help as I have never used one.

yes that where i got this
Delta T = (1 - (Heat Load/Max Cooling Power)) * Max Temp Difference

I also had to figure out how many watts i would get off of 12 volts vers 15V.

I just need someone to tell me my math is right.

been waiting who here knows the math?

Don't know if I'm right but I don't think you'll ever fit 24 - 26 amps through those wires or your CPU unless yer using like 00 or 0000 cable lol. My TEC I can't remember the numbers exactly, but was 176 Watts (I think) and was only rated for 12V and just over 2 Amps.

So not sure about your math but the current doesn't seem to be adding up.

EDIT:
Checked my email and those TECs were 136 Watts. But 12V and 9 Amps. And Molex, atleast to my understanding only puts out 5 Amps worth. And also, putting my retardation aside, ya the CPU can take like 75A which is amazing to me. So forgive my ignorance lol. Ya you should be able to fit 26 or so amps through the cables, but they would have to be upgraded to something more sturdy, but then again, I read the Molex only puts out 5 Amps so, unless that's wrong, don't think you need to worry about upgrading wires.

I did some math, and to get 300 Watts you will need 25 Amps on a 12V rail. if I am assuming every 12V to GND (Yellow to Black) is equal to 5A, you will need 2 6Pin PCIe power connectors. 3 From One connector, and 2 from the other.

According to my math, and a few assumptions about the Theoretical Max Temp Difference I got:

(LOAD) Delta T = (1 - (236Watts / 400)) * 67 = 27.5*C Difference (Hot to Cold)

(IDLE) Delta T = (1 - (12Watts / 400)) * 67 = 65*C Difference (Hot to Cold)

These are assuming that we're going to be running these TECs at 400Watts all the time (apparently that's very inefficient, even for TEC's)

(LOAD) Delta T = (1 - (236Watts / 300)) * 67 = 14.3*C Difference (Hot to Cold)

(IDLE) Delta T = (1 - (12Watts / 300)) * 67 = 64.3*C Difference (Hot to Cold)

In total, you're watercooling loop will have to be capable of removing about 540 Watts of power for a 100% CPU Load. I hope this is all the information you were looking for. Now finding out what is needed to remove that much heat, id another complex-ish problem all on it's own.

In addition to this, to find your CPU Temp with a sensor that does not work < 18*C or < 21*C like mine, all you have to do is subtract the temp of the liquid in your loop (a little redundant if you have a CPU Temp Sensor). So in my case,

(IDLE) Delta T = (1 - (24 Watts / 60 Watts)) * 67 = 40.2*C - 27*C = 13.2*C (+/-4*C because I'm taking a guess at my water temp)

It's been so long since I've thought of Parallel circuits, but could someone correct me if I'm wrong. If you take 2 sets of wires ([Yellow(12V,5A) and Black] x 2) and merge them into a single set of wires, would you have 12V and 10A, or Possibly 24V and 5A, or even 24V and 10A. I have a feeling it's the (12V, 10A).

This is the diagram I'm thinking of what's going to happen. Please correct it if it's wrong, or atleast give feedback: