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Recently I tried to cool water with one TEC1-12706 (waterblock+tec+cooler) but for a lots of reasons (that I am now aware) it didn't cool water (1L in a closed cicle) under or even near 0ºC. I rethinked and made some calculus and I concluded that maybe 4 TEC1-12708 should be enough.

The problem is that peltiers need to be well cooled, so I started to build another water refrigerator (bigger waterblock+4 peltiers+aluminium to dissipate heat+fans) but I realized that is a way more expensive (and i'm student) than buy the sets already made.

My question is if this set that I'm thinking to buy is sufficient to dissipate the heat of the hot side of the 4 peltiers. Peltier set image: https://imgur.com/sEw4UiN (assuming that cool side is turned to wateblock).

If not sufficient what are the other cheap ways to dissipate heat?

Thank you and sorry about my english.

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  • \$\begingroup\$ Read the datasheet of the TECs how much they dissipate, how much your cooling setup can dissipate, add some margni. \$\endgroup\$ – PlasmaHH Sep 19 '18 at 12:50
  • \$\begingroup\$ @PlasmaHH The first build I made, I made with a cooler (CPU cooler) that was made to dissipate 125w. But there are 2 problems: 4 CPU coolers are expensive, and how can I know how much heat can that set on the image dissipate? \$\endgroup\$ – João Raimundo Sep 19 '18 at 13:00
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Do you need small size or closed loop? If not, there are far better cheaper ways to provide cold water to your tec hot sides for cooling than a water block with fans.

A bong cooler can chill water to below room temperature, so gets you off to a good start with getting those hot sides cold. It's quite cheap to build one as well, one fan, and a pipe or container that can probably be scavanged or repurposed.

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  • \$\begingroup\$ Tthank you! I didn't know about bong coolers, they seem useful, but the idea is to be small cooler like 30cm*30cm*30cm "box" and maybe it is a small volume to make a bong cooler since I will need another water loop (?). Anyway I will search more about bong coolers \$\endgroup\$ – João Raimundo Sep 19 '18 at 13:10

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