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I made a cooling unit with two peltier(12706) with hot side attached to heatsink with fan. And cold side with waterblock i.e (heatsink + peltier(hot side with heatsink) then waterblock then another peltier (cold side with waterblock) then another peltier + heatsink) When I power the unit the waterblock get cold but what I run water through waterblock it doesn't cool . Why???

enter image description here

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    \$\begingroup\$ I suspect the Peltier module doesn't have enough heat exchanging capacity to cool the water at the rate you're pumping it. \$\endgroup\$
    – tangrs
    Commented May 3, 2018 at 4:59
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    \$\begingroup\$ Not sure if these images are your real unit, but don’t see any trace of a thermal compound for heat teansfer, however I also agree with @tangrs \$\endgroup\$
    – Solar Mike
    Commented May 3, 2018 at 5:07
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    \$\begingroup\$ How much decimals does your thermometer have? I'm sure it does change temperature, but it's by far not what you are expecting. \$\endgroup\$
    – Jeroen3
    Commented May 3, 2018 at 5:33
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    \$\begingroup\$ Are you trying to cool a continuous flow of water from an external source? Or are you trying to cool water that's circulating in a loop? How much water are you trying to cool? How much cooler are you expecting it to get? What's the flow rate? What thermal coupling do you have between the heat sinks and the block? \$\endgroup\$ Commented May 3, 2018 at 6:14
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    \$\begingroup\$ How much power, how much water, what Peltier? \$\endgroup\$ Commented May 3, 2018 at 6:29

2 Answers 2

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Assuming that your 12706 Peltiers are the same spec as this 12706 Peltier then.

The specifications are :

Imax: 6.4A
Umax: 15.4V
Qmax: (ΔT =0) 63 W ΔT max. = 68°C

To operate two at full power you need 15.4V at 12.8A or about 200 W of power.
Qmax = 126 Watts of cooling.
Assume say 100 W of available cooling.

Water will transfer about 4.2 J/K/cc Your 100W of cooling will cool 100/4.2 = 23.8 cc.K of water per second.
Lets say 25 cc.
ie cc x deltaT ~= 25 per second.

So 25cc per escond by 1 degree K or
1 cc per second by 25 degrees K or 0.5CC per second by 50 degrees K.
That doesn't sound much, but 0.5cc / second = 1.8 litre/hour x 50C drop.
Useful.

As dTmax = 68 C you cannot get more cooling by lowering water flow to much less than 0.5 CC/second.

The cooling is relative to the hot side.
To get cooling to under ambient the hot side should be as little above ambient as possible = lots of air flow.

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Added:

Do I really need thermal paste

Thermal paste is a VERY good idea.
Without paste you waste cooling by creating temperature differentials across the joints.

I have the same peltier , and I am running it on a power supply of 45 amp but on 12 volt( it's a computer power supply). Is voltage any problem. 15 volt will not affect them as everyone said they are 12 volt operational.

See datasheet. The 45A figure is the current your supply CAN provide at 12V - NOT what the load actually takes at 12V.

Also I am running it for one hour but still no effect on temp I think cooling should be slow but there should be some on 12 volt ,do you think it is totally useless on 12 volt

See datasheet.
Each Peltier takes about 6A at 15V and 4.5A at 12V.
The charts at the bottom of page 2 show surprisingly little difference at 4.5A or 6A.

Peltier Datasheet here

At 12V each device should draw 4 to 4.5A, and provide say 30W to 40W of cooling at 20 C differential. That's 70W for 2 or about 70% of what I assumed in my answer. Reduce flow rates to 70% of those above for same temperature drop. ie IF you get cooling with no water flow then a very slow water flow should be cooled. About 0.4 CC/second should give 25C cooling.

See datasheet - note that as Thot-Tcold differential rises the cooling Watts drop.

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  • \$\begingroup\$ I have the same peltier , and I am running it on a power supply of 45 amp but on 12 volt( it's a computer power supply). Is voltage any problem. 15 volt will not affect them as everyone said they are 12 volt operational. Also I am running it for one hour but still no effect on temp I think cooling should be slow but there should be some on 12 volt ,do you think it is totally useless on 12 volt \$\endgroup\$
    – Umar Usman
    Commented May 3, 2018 at 16:42
  • \$\begingroup\$ Do I really need thermal paste \$\endgroup\$
    – Umar Usman
    Commented May 3, 2018 at 16:59
  • \$\begingroup\$ @UmarUsman Without paste you waste cooling by creating temperature differentials across the joints. Thermal paste is a VERY good idea. | Peltier Datasheet here . At 12V each device should draw 4 to 4.5A, and provide say 35W of cooling at 20 C differential. That's 70W for 2 or about 70% of what I assumed in my answer. Reduce water flow rates per temperature drop to 70% of above for same cooling. ie IF you get cooling with no water flow then a very slow water flow should be cooled. ... \$\endgroup\$
    – Russell McMahon
    Commented May 6, 2018 at 12:58
  • \$\begingroup\$ @UmarUsman ... About 0.4 CC/second should give 25C cooling. If you use much higher flow rates the cooling may be too small to sense. \$\endgroup\$
    – Russell McMahon
    Commented May 6, 2018 at 12:59
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It sounds like you put the system together backwards, usually you put the peltier element in contact with your heat source and (water)cool the hot side. If you require the cooling of 2 units at once I suggest a copper plate(about 10mm thick) to spread the heat to both and cool the hot sides as you se fit. Do not forget that this (and any non passive cooling) can cause condensation if not insulated correctly.

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