Lehpron From United States of America, joined Jul 2001, 7028 posts, RR: 22 Posted (6 years 7 months 2 weeks 23 hours ago) and read 1901 times:
I'm planning to use a TEC (thermo-electric cooler) in my computer and the math of these things are throwing me off.
The simplest and most inexpensive approach I was thinking of was using a small 30watt TEC on the north-bridge of my motherboard using the stock heat sink. I did some analysis but I am confused as to how much heat is coming out. The TEC device I was planning on is by a company in New Jersy called Melcor, here's the link for the device.
These are the important stats:
Vmax=15.4 volts
Imax=3.9 amps (max current draw at maximum voltage)
Qmax=33.4 watts (at maximum current and voltage)
Note: according to Melcor, Qmax is the maximum heat absorption, the cold side.
Using the equation V=IR,
Resistance of the device at Vmax/Imax is just under 4 ohms.
Assuming that R is constant with temp, this TEC draws out a bit over 3 amps @ 12.38 volts from my PSU, which is 40 watts of electricity to run.
I've set up an efficiency rating, that ratio of Q/(IV) should show how much Qmax is expected per total power used. For this device that efficiency is 0.55611 @ 25*C. If Melcor intends Qmax to be the coldside, that implies a 55.6% efficiency...is that realistic? That would mean that 44.4% of power going in comes out as heat. It is hard for me to accept that which is why I feel I may have made a mistake.
As of my north-bridge, I'm not worried. The power levels are small and the stock aluminum heat-sink can easily pump out 22 watts of heat, if I think Qmax is the heat expelled to run the device.
Ideas, anyone?
The meaning of life is curiosity; we were put on this planet to explore opportunities.
StrasserB From Singapore, joined Nov 2005, 1541 posts, RR: 8 Reply 1, posted (6 years 7 months 2 weeks 23 hours ago) and read 1893 times:
To be honest, it might be that I didn't understand all of this technical stuff; but just one question: Are you sure, that the mentioned north-bridge is correctly located at 32°49'28.10" N 117°10'51.30" W?
Still, even in the most arid desert is an airport somewhere ...
Lehpron From United States of America, joined Jul 2001, 7028 posts, RR: 22 Reply 2, posted (6 years 7 months 2 weeks 23 hours ago) and read 1884 times:
Forgot to mention, I want to figure out this so I could try something a bit more complex, like adding a 226-watt device to my videocard...which is a risk for now.
For those who may not know, the north-bridge is usually located next to the main cpu on the computer desktop motherboard, it is the 'hub' of the motherboard. It properly allocates information from and to the major parts like the processor, memory, hard drive, etc. It gets hottest (for me) during 2D applications, like internet (right now it is about 51*C or 124-degrees F)
The meaning of life is curiosity; we were put on this planet to explore opportunities.
Ilikeyyc From United States of America, joined Dec 2003, 1373 posts, RR: 22 Reply 3, posted (6 years 7 months 2 weeks 22 hours ago) and read 1877 times:
unless I am missing a brain cell or two:
%eff= 100 x Q/VI
Q= 33.4
V= 12.38
I = 3.2 (you estimated just over 3, so I used this number)
Lehpron From United States of America, joined Jul 2001, 7028 posts, RR: 22 Reply 4, posted (6 years 7 months 2 weeks 20 hours ago) and read 1863 times:
Quoting Ilikeyyc (Reply 3): Q= 33.4
V= 12.38
I = 3.2 (you estimated just over 3, so I used this number)
33.4/(12.38 x 3.2)=
33.4/39.616=
.8431, or 84.31% efficient
The Qmax of this device only occurs when the maximum voltage and maximum current is applied, since my voltage os lower and the coresponding current is lower, the Q value should be less than max, certainly cannot pass whatever comes. If the voltage was 6v the power requirements wou be 20watts, which the 33watts is greater than 100% efficiency. I still don't know if the values are either heating or cooling.
The meaning of life is curiosity; we were put on this planet to explore opportunities.
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