# I want to dissipate 3 kW of heat with an air conditioner. How big of an air conditioner do I need?

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, the 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?

• BTU can be converted to Watts per hour, so do that. Apr 5, 2021 at 6:22
• See, after a quick Google search: unitconverters.net/power/watt-to-btu-it-hour.htm Apr 5, 2021 at 7:21
• @SolarMike: "watts per hour"? Did you mean "watt-hours"? Apr 5, 2021 at 11:04
• so Bitcoin mining. Is this possibly an XY problem? Do you really need an air conditioner or can you put your miners outside where it's colder? Sep 3, 2021 at 16:17
• It seems to me you are maintaining the room temperature around 60 degrees C, if outdoor is such cold, how? Do you keep heating on? You need to maintain the temperature below 60 degrees, how far below?
– r13
Sep 28, 2021 at 16:30

You can either:

• strive to have the temperature inside lower than outside and use AC to achieve that. In that case you want your windows to be closed and the walls to be insulated as much as possible. To maintain low temperature inside you need an AC unit capable to dump the 3kW of heat generated by the devices, as well as any extra heat coming inside from the (warmer) outside through (not perfectly insulated) walls/windows and the solar heat. You need the AC unit to exceed these requirements by some margin. Otherwise, even a slight excess of heat not dumped by the AC will result in temperature rise to the point that it becomes warmer inside than outside, at which point, you are better opening your windows/doors, and the walls' insulation is working against you. Having a low temperature inside obviously allows you to achieve your goals of keeping the devices at 60C or below. Additionally, the lower the temperature inside, the slower the devices' fans need to work, and thus they become quieter and consume less energy. Or,

• you can allow the temperature inside to be higher than outside. With this approach you want as little insulation on your walls/roof as possible (maybe only use some reflective material, or white paint, to fend off solar radiation), and you want the window open. Possibly you can use fans to actively push/pull the air through the window. You can either use an AC, or not. But even if you do, you do not use it to get lower temperature than outside, but simply lower than it is currently. Obviously, the higher the temperature you allow inside, the easier it is to maintain it. At some point no AC is needed, and simply venting through the window and through the (not isolated/not well isolated) walls and roof is enough. Even with closed window the temperature won't rise above some threshold, because the heat from the hot room will leak outside through the walls/roof. Actually, the hotter the room the more heat it dumps by its own through the walls. However, the hotter it gets inside, the devices' fans need to work harder to maintain the required temperature of 60C. At some point, if the room gets too hot, the devices' fans may reach their top speed and not be able to work any harder. You can simply replace the fans with larger/stronger ones, but obviously no fan will be able to cool the devices/heat sinks if the room temperature gets above 60C. If you keep your room hotter than outside you can save on AC, but your devices' fans work harder and consume more energy, and any fans that push/pull air through the window do so as well. Still, as pointed out by NMech and Transistor in their answers, it probably takes less energy to have these devices' fans (plus additional fans by the window) spin like crazy, than to cool the entire room to very low temperature so that these fans do not need to spin so fast.

One caveat is that not all computing devices like to work in high ambient temperatures. You have not specified what your devices are, but likely they are some servers/mining rigs. Even if the CPU can safely get to 60C, other components not necessarily so. E.g. Dell warranties their servers to operate at up to 45C (also at this temperature the servers' fans should be able to keep the right temperature internally).

I think it is safe to keep the room at 45C, but when it gets hotter, you are better of switching your devices off, or running some AC to alleviate the situation. Chances are that with temperature outside in the range 20C-35C you will never have to run AC.

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.

• Why is it only a "type" of heat pump? Apr 5, 2021 at 11:59
• @SolarMike, maybe "class of heat pump" would be better. Apr 5, 2021 at 12:03
• @Transistor, it's fine Apr 5, 2021 at 13:31
• 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 Apr 6, 2021 at 0:51
• $\Delta T$ is the temperature rise of the air. Apr 6, 2021 at 8:59

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.

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)
• I'm genuinely interested as well! (I have upvoted) Sep 28, 2021 at 14:13
• @r13 Interesting... I understood it differently: the fact that 60 degrees is achievable now, does not mean it is barely achievable and any increase in the outside temperature would cause the 60 degrees to be exceeded. Besides the 60 degrees is IMO the temperature of the devices, not of the room. Sep 28, 2021 at 16:31
• @r13 I can barely see a reason to insulate a tiny room which constantly generates 3kW of heat. Even in cold climate. There would be never need to warm it up, or conserve heat, on the contrary, there would be constant need to get rid of the excess heat. The best you could do is to remove insulation, or actually completely remove the walls (but then it would be unprotected against rain/thieves etc). Sep 28, 2021 at 18:29
• @r13 If it is hotter outside than inside, then ok, it makes sense to insulate the walls. I guess it can only be colder inside if we run AC like crazy, to not only dump the excess 3KW of heat, but also get below ambient temperature. On the other hand, if the target temperature is 60 degrees C, then it makes really no sense to cool the inside so much as to have a lower temperature than outside. The OP is talking about opening a window - it can only make sense if inside is hotter than outside. Sep 28, 2021 at 18:48
• @r13 Basically, if outside is 20C and inside is 40C, and I open the window, does it draw more heat in? or does it leak the heat out? Sep 28, 2021 at 21:53