# Playing with models: quantitative exploration of life.

## Debunking the “Keep the AC on while away” myth

Posted in Economics by Alexander Lobkovsky Meitiv on September 22, 2015

We all heard the supposedly common wisdom of leaving the AC (or heat) on in the house when away on a trip. This “wisdom” is perpetuated by the HVAC professionals themselves. Well, it’s wrong, at least under the simplifying assumptions below. You will always save energy and money by turning the heat/AC off when you go on a trip, no matter how short.

To quantitatively compute the difference in the money/electricity for the two scenarios: 1) maintain house temperature, and 2) turn the heat/AC off while away we will need to know the cost of either removing or adding the amount of heat $Q$ from/to the house. I will assume that the cost is simply proportional to $Q$. This is probably a good assumption for combustion heating systems but may not be very accurate for heat pumps whose efficiency depends on the temperature difference between the inside and outside. Let me deal with this complication later.

Let me assume that the house has heat capacity $C$ so that the amount of heat $Q$ that enters or leaves the house to cause the temperature change $\Delta T$ is $Q = C \Delta T.$ Let me also assume that total heat conductance of the house’s walls is $K$. This heat conductance is the coefficient of proportionality between the outside-inside temperature difference and the total heat that flows into/out of the house per unit time: $dQ/dt = -K \Delta T.$

Therefore, under scenario 1 (keep heat/AC on), if the duration of the trip is $\tau$, the total heat that needs to be pumped into/out of the house is $Q_1 = K\tau |\Delta T|$, where $|\Delta T|$ is the mean difference between the indoor and outdoor temperatures.

When heat/AC is off, the house temperature will approach the mean outdoor temperature exponentially with the charateristic time scale $C/K$. Let’s assume that the heat/AC system is so powerful that the heat is removed/added quickly when you arrive from the trip so that the extra losses during the cooling/heating period are small compared to the total amount of heat to be removed/added. Then, the total amount of heat to be removed/added after the trip is $Q_2 = (1 - e^{-K\tau/C}) C |\Delta T|.$

Comparing the two amounts of heat we see that because $x > 1 - e^{-x}$,

$Q_1 > Q_2,$

In other words, the amount of heat that must be removed/added from/to the house is always greater when you keep the system on during the trip. Therefore, unless there are counter indications like the pets dying or pipes freezing, it is better to turn the heat/AC off when you travel or at least change the thermostat to a setting that is closer to the outdoor temperature.