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So the title says it all.

I have a room (4 m x 4 m x 2.5 m) which is closed.(Actually there is a window and I could add a 50 cm panel fan if needed as an exhaust if that will help.)

I have some devices that need to be under 60°C they manage to do that now but it is still cold here. (Average temperature outside is about 5 to 15°C ± 3°C and I leave the window open, they devices are actively cooled with heat sinks and small fans that work up to 50 - 80% of their speed.)

The total power consumption of the devices is about 3000 watts.

How big of an air conditioner would I need to maintain about the same thermals if the outside temperature gets to like 20 to 35 degrees outside?

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  • $\begingroup$ BTU can be converted to Watts per hour, so do that. $\endgroup$
    – Solar Mike
    Apr 5 at 6:22
  • $\begingroup$ See, after a quick Google search: unitconverters.net/power/watt-to-btu-it-hour.htm $\endgroup$
    – Solar Mike
    Apr 5 at 7:21
  • $\begingroup$ @SolarMike: "watts per hour"? Did you mean "watt-hours"? $\endgroup$
    – Transistor
    Apr 5 at 11:04
  • $\begingroup$ @Fred, tip: You can also use HTML entities &Omega;, &mu;, &deg;, &times;, &pm; etc. as well as <sup>...</sup> and <sub>...</sub> in the posts (but they don't work in the comments). $\endgroup$
    – Transistor
    Apr 5 at 11:45
  • $\begingroup$ @Transistor: Thanks for the tip. I'm an old dog happy to learn new tricks (tips) ;-) $\endgroup$
    – Fred
    Apr 5 at 17:19
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Air conditioning systems are a type of heat pump. Heat pumps are used to remove heat from a system at one temperature and transfer it to a system at a higher temperature. The heat pump lifts the energy from the lower temperature to the higher.

In your case the outside ambient temperature appears to be always much lower than the required operating temperature so there is no need for a heat pump (or air conditioning). You should be able to achieve adequate cooling by air circulation.

As a rough guide to the required airflow we can do a quick calculation. The specific heat capacity of air is approximately 1 kJ.kg-1.K-1. If you want to limit your air temperature rise to 10°C then we can calculate as follows:

$$ t = \frac {\Delta T \cdot m \cdot SHC} P $$ where $ t $ is the time, $ \Delta T $ is the temperature change, $ m $ is the mass of air, $ SHC $ is the specific heat capacity and $ P $ is the power. Rearranging we get $$ \frac m t = \frac P {\Delta T \times SHC} $$ and for 3 kW of cooling with 10°C air temperature rise we get $$ \frac m t = \frac 3 {10 \times 1} = 0.3 \ \mathrm {kg/s} $$

Air has a density of 1.225 kg/m3 at sea level so your required airflow is $ \frac {0.3 } {1.225} = 0.25\ \mathrm {m^3/s} $.

Bear in mind that even if you use air conditioning that you'll still have the problem of moving the air around the room to ensure elimination of local hotspots.

This simple fan solution should be much simpler, lower cost and environmentally friendly than an air conditioning system.

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  • $\begingroup$ Why is it only a "type" of heat pump? $\endgroup$
    – Solar Mike
    Apr 5 at 11:59
  • $\begingroup$ @SolarMike, maybe "class of heat pump" would be better. $\endgroup$
    – Transistor
    Apr 5 at 12:03
  • $\begingroup$ @Transistor, it's fine $\endgroup$
    – Tiger Guy
    Apr 5 at 13:31
  • $\begingroup$ Thanks @Transistor but I dont get something in this formula, if we assume temperature delta of 20 (so from an environment of 10C to an environment of 30C) or is the temperature delta between the temperature difference of the device and the environment? either way the bigger the delta the less the airflow I need? ok typing it down it makes sense since I realize it about the delta between the device and the environment bigger delta is (relative to the device) colder environment. so I dont need to post this but I still gonna because I wasted so much time typinghahaha $\endgroup$
    – papajo
    Apr 6 at 0:51
  • $\begingroup$ $ \Delta T $ is the temperature rise of the air. $\endgroup$
    – Transistor
    Apr 6 at 8:59
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One simple and effective way is to install an intake window with adjustable louvers near the lower part of your heat sink and a hood (can be possibly a primitive cardboard one) on top leading to the window/ vent on top.

This way you create a self-sustaining circulation system that takes advantage of the rising tendency of the hot air coming off the heat sinks.

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You probably don't need any airconditioner. if the temperature outside remains a constant $20-25[^oC]$ then the heat transfer through the walls should cool the room down under $60[^oC]$.

The only reason that might not be if you leave in a country with significant sunshine, and instead of having a good insulation at the roof you have something that heats up.

Instead of an airconditioner, (if you are not concerned about being at 40C, or humidity) I would just suggest that you do a half decent heat instulation at the root and leave it at that.


If you insist on the airconditioner, you need to provide information about the wall heat conductivity coefficients and thicknesses.

Just go give you a very rough indication. Assuming that:

  • all walls are exposed to $25^oC$,
  • the thickness of the wall is 10 [cm]
  • the wall is made out of fire brick (which does not propagate heat that much) with $k = 0.5 [J/mK]$

Then the overall temperature losses from all the side walls and the roof for a uniform temperature of 55[C] will be about 8.4kW.

For the 3[kW], you should expect a (uniform ) temperature difference of about $10 [K]$, i.e. the room temperature will be about $35[^oC]$.

IMHO, you could optionally (even that I think is an overkill):

  • add louvres (top and bottom of the side walls) (As kamran suggested)
  • add a fan inside the room pointed at the heat source (that will cool better and distribute the temperature in the room)
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  • $\begingroup$ Because, this my most downvoted answer - so far- , It would mean a lot for my personal improvement if people that found this answer was not useful, took a moment to comment on why they feel so. $\endgroup$
    – NMech
    Sep 3 at 7:26

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