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I want to know is possible cooling a box in 1 * 2 m size by peltier? please help me thanks

رحمن محمدی
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  • Yes it is possible. How much cooling you can get will 'vary', and might be quite insignificant. – Spehro Pefhany Oct 03 '20 at 05:32
  • To calculate this properly your going to need to know the third dimension of the box, the required change in temperature, the U-vale of the insulation, the heat capacity of the box + its contents and the cooling rate or time. From that you can calculate the cooling power required and then you can start selecting your Peltier modules. – Transistor Oct 03 '20 at 05:51
  • if add 2 peltier or more, can it help to cool more? – رحمن محمدی Oct 03 '20 at 05:57
  • 1x2 metres, no, think compression cycle (normal fridge concept) cooling. Pelttiers work well in 5 litre beer-coolers. – Neil_UK Oct 03 '20 at 06:05
  • "if add 2 peltier or more, can it help to cool more?" Possibly, but you still have no specifications. – Transistor Oct 03 '20 at 06:16
  • Suppose we want to cool a wall rack whose switch heats up to 60 degrees to 18 degrees. Can a plotter help in this case? – رحمن محمدی Oct 03 '20 at 06:27
  • (1) What is a "wall rack" and what is its switch? (2) Is this an electronics cabinet? (3) Degrees C or F? (4) what electrical power is being consumed by the equipment. Please [edit] your question to supply ^^all ** the missing information. – Transistor Oct 03 '20 at 07:56
  • Tnx a lot for reply,1-2 its a network wallmount rack with a network switch.3-C & for Q 4 the temp of switch about 60C but i dont know about "what electrical power is being consumed" – رحمن محمدی Oct 03 '20 at 08:54
  • You'll need to measure it. Otherwise you are guessing. Why would a cabinet fan not provide adequate cooling? What is the ambient temperature and what temperature do you want in the cabinet? Don't forget to [edit] all this information into your question. – Transistor Oct 03 '20 at 08:57
  • Sorry Transistor i cant edit my question. The rack is at the top of the shed and the shed does not have proper ventilation, so the air above the shed is very hot. I was able to lower it up to 10 degrees with the ventilation fan, but the switch temperature is still high – رحمن محمدی Oct 03 '20 at 09:38
  • To edit your question click the "edit" tab under the question. Please add information that you have provided in comments. | It's essential to now wattage of object to be cooled. – Russell McMahon Oct 03 '20 at 10:17
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    There is an [edit] link under your question and I have provided it twice in the previous comments and again here. You're asking for an engineering solution to a problem. That means we need numbers. "Very hot" is not a temperature measurement. "Still high" is not a temperature measurement. You need to measure the power input if you are to calculate the cooling required. – Transistor Oct 03 '20 at 11:20
  • With 2 Peltier modules, you'll have to invest a lot in sensitive temperature monitoring to detect any cooling at all. –  Oct 03 '20 at 13:40

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I assume your box has a height of 1m, thus it could store 2m³ of water. I fill it with only 500L. And the water has a temperature of 25°C. I'd like to drink it cool and I prefer 15°C. This makes a difference of 10K. To remove 10K of warmth from 500L of water (4 kJ/(kg*K)) requires to extract 20MJ of heat.

This will not happen instantly like switching the peltiers on and 20MJ are gone. Instead this needs some time, I assume a day or 24h. Dividing 20MJ with 24h gives 834kJ/h. And kj/h can be converted to kW which is roughly 0.232kW or 232W for an entire day.

Finally peltiers aren't that efficient and if you load your set of peltiers with 250W they won't extract the same amount of heat. This largely depends on the thermal resistance on the hot side. A flat peltier in the desert sun experiences a high thermal resistance causing very low efficiency. If you add instead a good heat sink and maybe a fan, too you'll end up in the 20% range, thus 1W of electric power turns into 0.2W of extracted temperature power. (And also add the fan's power consumption and cost for the heat sinks. The box would also be dug into the ground in the shadows.)

Now this is where the guessing starts and I multiply 0.232kW by 5 (for the very optimistic 20% efficiency of your overall construction cooling 500L of water from 25°C to 15°C within 24h) and this yields 1.25kW of electric power each hour for an entire day. With 5V elements in parallel you'd have to supply 250A to all of them.

If you need the formulas feel free to comment, so I add it later on. I also used common »power« instead of distinguishing between scientific power and energy.

motzmann
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