Originally Posted by dizzy4
Basically I am ill-informed about using TECs and calculating their properties. This is made easier by ultrasonic2's calculator (Thanks for that), but I cannot put in specific values and must use the pre-set ones. I also have a hard time reading the charts and determining the voltage to use on the TEC. This is important because I need to know which meanwell psu to get. I do understand about deltaT and efficiencies, but I am less informed about how Qmax and some other things factor in.
As always, any help is appreciated!
Firstly as your using a single TEC it is easiest to just use a 15.6v TEC at 12v this will give you close to what many consider the "ideal operating point" of 70-75% power. This will in the main get rid of most of the concerns pressing you.
If you want to read the charts it is really quite easy once you know how.
The best charts to start with are ferrotec. click on a TEC code on one of these 2 pages to see some - http://www.ferrotec.co.uk/products/t...s/singleStage/
or the higher power tec's at - http://www.ferrotec.co.uk/products/t...les/highPower/
Ferrotec's naming is simple ignore the first 4 numbers the next 3 after the hash are the number of junctions. All Bismuth telluride TEC's used for cooling have the same "make up" and you multiply the number of junctions by 0.123 to get the nominal voltage at 25ÂºC so 127 is 127*0.123 = 15.6v at 25ÂºC. The 3 numbers after the next hash are the current rating which you get by dividing this number by 100 so 060 is = to 6amps and 085 is = 8.5amps.
All TEC's with same no. of junctions have basically the same characteristics so to get a ballpark figure you can use any 127 junction chart with a close current rating and just put the Vmax and the Imax of the TEC you want to use on the axis instead of those already there.
For example go to http://www.ferrotec.co.uk/products/t...s/singleStage/
and click the 9500/127/100 B somewhere near the bottom.
It will show 4 charts but Ferrotec use the default hotside temp of 50ÂºC, just under the charts is a box says "Hot side temperature." change this to a more reasonable value if you are going to have good quality water cooling on the hot side a good ballpark figure to start from is 10ÂºC ABOVE ambient, so add this in the box RECALCULATE and the charts change.
FOR THIS EXAMPLE SET THE TEMP TO 30ÂºC.
This TEC has a qcmax of 95w at 50ÂºC at full power and has Umax of 17.5v and Imax 10amps and we will calculate the specs for it when the hotside is at a temperature of 30ÂºC
The first chart
is voltage V current on the top right click on it to enlarge. Select the voltage you want to use on the vertical axis and read across... the lines are the temperature deltas i.e. the difference between the hot and cold sides. After discussions with Uncle Jimbo on another forum we decided that it would be fair, dependent on the quality of your water cooling you could assume a dt of between 10Âº and 20Âº. I must point here it is only a BALLPARK figure we are going to get...just something you can start from.
So for now find 12v on the vertical axis draw a line across from the voltage to a point between the blue and green lines then trace vertically down to get the current draw at 12v at 30ÂºC - it is 7.75amps (7.5 plus one notch.) remember these are rough figures. Make a note of the current.
The second chart
is heat pumped v current on the top left hand side.
Click to enlarge. find 7.75 amps on this charts horizontal axis it is the same 7.5 plus one notch. Then trace up vertically to between the blue and green lines again then trace to the left to get the Qc (heatpump) it is around about 60w. remember this value is only valid at the hotside temp, voltage and everything else you have used.
The third chart
on the bottom left is what everyone wants to know - how much cooling you have to apply. This figure is the pump calculated with the previous chart plus the heat generated by the TEC.
Again just find the current 7.75 amps you noted from the first chart on the horizontal axis and read up to between the blue and green lines then trace left to get Qh (total heat rejected by the TEC.) It is around about 160-170w. This figure is what you have to cool with your rad.
The fourth chart
on the bottom right is the coefficient of performance. A COP of 1 indicates you are using as much power as you are cooling by. A figure less than 1 indicates you are using more power than you are cooling by i.e. inefficient. So obviously a figure bigger than 1 is an efficient system. A COP of 2 indicates the power of your cooling is twice the power you putting in to the unit. You probably guessed by now you just look up 7.75 on the horizontal axis and read up to between the blue and green lines then trace left to get the COP. It is around about 0.75...ouch that's not very good is it.
So to recap with good watercooling when run at a hotside temp. of 30ÂºC using a voltage of 12v it draws 7.75 amps for a pump of about 60w. It runs at a COP of 0.75 and will need cooling to the tune of 160-170w.
EASY ISN'T IT ???
well it is with the ferrotec's other charts take a bit more brain power.
You must remember though the figures you arrived at are only valid at the voltage/current and hotside temp. values you initially used. The problem with TEC's is every figure depends on something else and the ONLY static spec. of a TEC is it's physical size.
If you click on the advanced settings under the charts and you know the thermal resistance ( ÂºC/Watt ) of the heatsink you plan to place on the hotside of the TEC the charts will alter to take this in to account.