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I'm looking for a way to keep a liquid under 50°C. This liquid has a thermal conductivity of .1396 and a specific heat of 2.054 both at 40°C. and a constant energy source submerged in it ranging from 16 kW to 17.5 kW.

I'm barely starting to read about this topic, according some forums, a car generally transfer 1/3 of the engine power through heat to the coolant that gets circulated, this would mean that even for a small car engine ( say a 90's corolla ) the radiator would have to dissipate at most 24 kW, assuming that the engine produces about 100 hp.

My questions are, is the previous scenario correct? How do you calculate the amount of heat any particular radiator can dissipate? Would an liquid-to-air heat exchanger be the right solution for my original problem? If not what would suggest?

Sorry for the vague question, but hopefully someone can help out and point me in the right direction. Thanks!

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    $\begingroup$ Why not evaluate the amount of heat to be removed in the worst condition, then size a heat exchanger to match. The are several good texts about heat exchangers that will give you the theory and you will get an idea of size. Then any heat exchanger manufacturer can check your calculations and supply the correct one. $\endgroup$
    – Solar Mike
    Aug 9 '21 at 8:27
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    $\begingroup$ The units for the thermal conductivity and specific heat values you mention should be included in the question. $\endgroup$
    – J. Ari
    Aug 9 '21 at 11:27
  • $\begingroup$ You will also need to know the air temp you are using to do the cooling. This is much easier in Antarctica in winter than in Death Valley in the summer. $\endgroup$
    – Tiger Guy
    Aug 10 '21 at 5:28

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