# Should the weaker of two air conditions be switched off?

We have two airconditioning systems in the office. For the sake of this question, let us assume the following current conditions:

• AC1 (old) outputs air of 26°C
• AC2 (new) outputs air of 18°C
• Office temperature is currently 23°C
• outside temperature is 35°C

Now as far as I know, air conditioning systems are a kind of heat pump; they transport energy through the temperature gradients $\text{office air }> \text{cooling coils}$ and $\text{outside coils} > \text{outside ambient}$.

Now obviously the first one doesn't hold for AC1, and as such I would say it is working in reverse: pumping energy from outside to inside the building. So for my intuition it would be better to switch off the old one until office temperature reaches 26°C, if the objective is the keep the office as cool as possible.

Is this intuition right and can it be backed up by a bit more physics calculations? My intuition says that it is the same as opening a window (and letting the warmer air in) or letting sunlight into the office by lifting the outside sunshades.

Note: I am ignoring the fact that we have a fresh air heat exchange system since that should in theory not bring any substantial energy inside.

• Do your A/C units also serve as dehumidifiers? And what type are they? Are they built into the wall? Or are they portable? Commented Jul 3, 2015 at 10:28
• @lemon: No idea about any specifics regarding dehumidification, but they do (humidity is rather low inside). By type do you mean the model number? I do not have it. The outlets for the cool air are built into the ceiling, connect via pipes to the units that disperse the heat on the top of the building. Commented Jul 3, 2015 at 10:44
• I'd be very surprised if such a situation were possible: the temperature of the evaporator coil ('cold' coil) in a normal air-con unit is approximately 5°C. If your room is at 23°C, there's no way the output of AC1 could be higher than 23°C. You may find that the air temperature in equilibrium with just AC1 running is 26°C, but this is simply a balance between the cooling produced by the AC and the heat 'leaking' into the office from outside. Commented Jul 3, 2015 at 12:51
• (As Chowie says in his answer, if the air coming from AC1 is genuinely at 26°C this suggests there's something very wrong with the unit!) Commented Jul 3, 2015 at 12:55
• @tok3rat0r: Yes, I think the old unit can't properly get rid of the heat on the top of the building because it is heated up by the sun there. The temp probes are directly inside the air outlet. I can not reach to the evporating coils since there is the fan in the way. I was hoping for some back of the envelope calculations that back this up, like "compressing the refrigerant adds roughly this energy, producing temperature differential of X between cold and warm, assuming warm is 26° this means outside is at least N° but to have it functional should be at most M°" or something like that. Commented Jul 3, 2015 at 13:02

Without knowing a lot more about the details of the entire system, we can't say exactly what the difference will be between the AC1+AC2 case and the AC2-only case. However, there are a few things we can work out from simple energy balance.

In the following, I'm considering the air as an ideal gas with a fixed specific heat capacity at constant pressure $c_p$ in the temperature range of interest. This means that no account is made of the different relative humidity of air coming from the two systems. Since the air coming from the AC units is pretty dry anyway, this should be a fairly good approximation.

We have some mass of air (internal to the office) passing through each of the air-con systems per unit time: denote the indoor mass flow rate through AC$i$ as $\dot{m}_i$. There is also a source of heat from outside the building (through the walls, windows, doors etc.) which we denote $Q_{th}$. This isn't known based on the details in the question, but a pretty reasonable assumption is that it's proportional to the temperature difference between the outside air and the inside air. Call it $\dot{Q}_{th}=C(T_\mathrm{out}-T_\mathrm{in})$, where $C$ is some 'thermal transfer coefficient'. Now, we know that in equilibrium we must have energy out = energy in. In terms of the figure, this equates to $$12C+3c_p\dot{m}_1-5c_p\dot{m}_2=0\:.$$ We wish to know how much lower the temperature in the office (currently 296 K) will be if we switch off the unit AC1. This is equivalent to setting the mass flow rate $\dot{m}_1$ to zero. We assume that, since the room was warmer than its set-point temperature, AC2 was already working flat out and so neither $\dot{m}_2$ nor the output temperature change. The equation for energy balance with our new internal temperature $T_{new}$ is $$(308-T_{new})C-(T_{new}-291)\dot{m}_2=0\:.$$ Equating these two expressions (since $0=0$), we can rearrange to get $$T_{new}=296-\frac{3c_p\dot{m}_1}{(C+c_p\dot{m}_2)}\:.$$ We know something else which allows us to eliminate one of the quantities in this expression: in the limit $C=0$, i.e. where the room is hermetically sealed, $T_{new}$ must equal the output temperature of AC2, i.e. 291 K. We can set $C=0$ in the above expression to find that $\dot{m}_1=5\dot{m}_2/3$. Plugging this in, we therefore have our final expression $$T_{new}=296-\frac{5c_p\dot{m}_2}{(C+c_p\dot{m}_2)}\:.$$ This expression has the behaviour we're looking for; it goes to 291 K when $C$ goes to zero, and as $C$ gets more significant relative to $c_p\dot{m}_2$ it approaches 296 K.

And that, I think, is about as far as we can go without knowing in full detail the thermal properties of the office such as its insulation, surface area, material composition and everything else!

If you're objective is to cool down the office, and you have a machine blowing air into the office which is warmer than the current (indoor) ambient temp. then I would by all means turn the damned thing off!

air conditioners are heat pumps, just like a fridge. They take in the ambient air, pass it over a cool surface (to absorb the energy from the air) and blow that (cooler) air back into the room. Through mechanical wizardry the excess heat is expelled (outside) via a 2nd fan.

Hence if your a/c is actually blowing warmer air than the room has in it, then I suggest you've either got it hooked up in reverse and you're actually trying to cool the world (admirable!) or the cooling stage has failed and it is no longer doing any cooling and is simply warming up the air with all the work it's doing.

Google for heat sinks and the three laws of thermodynamics if you'd like to understand the underlying physics of what is occurring and why the above is true.