This reminds situation reminds be of an episode from the The Big Bang Theory, although it is quite different from it. My roommate and I have similar temperature preferences and also share the objective of saving money on electricity. The optimal solution towards this objective is what we have differing opinions about.

The problem:
In my opinion, during the daytime when no-one's around, the temperature setting should be at least 4 °F more than the normal setting (given an ongoing summer 90+ °F). I would go far as to argue that during the daytime the cooling should be completely switched off. My roommate's counter-argument is that assuming a PID controlled model, the overshoot caused by the cooling in the evening, in an effort to get the temperature down from the day's high, will more than nullify the electricity saved during the day thereby resulting in higher costs. The controller is a programmable digital thermostat with 4 separate settings for sleep, wake, leave and return for two regimens namely weekday and weekend.

Instead of hand-waving I was looking for a mathematical model to simulate the heating-cooling of a typical apartment. A quick Google search yielded no useful models (most I could get my hands on are macro-models related to multiple commercial units).

I am looking for a model — with elements like the heat exchange (i.e. the ac), the heat sink (i.e. the apartment) and the environment — a general model which can be used for a first-order approximation. Assume that the cost of electricity is constant throughout the period under consideration (even if it isn't, a cost vs time graph can be easily incorporated in the model).


A number of factors go into determining the optimum daytime temperature settings for cooling: one of them is humidity. An air conditioner doesn't 'cool', it removes the heat, through the heat of the liquid refrigerant vented to the atmosphere. More humidity means in a residential setting without a lot of tonnage of air being moved is more 'wattage' is required to remove the 'moisture soaked' heat to bring down the temperature to a comfortable level(lets say 72F). Outdoor temperature and exposure to sunlight of course directly affect indoor temperatures.

Ideal cooling cycles at 3 times per hour, from 7 to 10 min. per cycle. This of course takes into consideration your cooling system is efficient(condenser coils cleaned, not blocked, filters clean, system sized correctly for your unit) and I'm assuming that you have a central air conditioner, and not a 'window shaker', which may not adequately move the volume of air required to cool. Depending on the outdoor temperature and humidity, it's usually a good idea to keep your unit at 78F when not occupied, this keeps the recoverability time shorter, as well as removing humidity and avalanching heat.

However, it may not be the optimal 'savings' you desire, as many utilities have off peak metering, meaning they give you a special rate on your electric bill if you install a 'saver switch' which turns off your a/c when the demand(usually commercial) is greater during the day, and then turns it on at 4:00-5:00PM when the commercial buildings go in to energy conservation. Since it sounds like you guys pay your energy bill, you should check into this option as it provides a guarenteed savings(whether they peak meter or not). Otherwise, 78F(or 80F if dryer climate) is a good daytime optimal setting.

In response to your comment about a 'model', there is a model in place, it's called a "Heat-Loss" calculation. This model is what is used to best size your heating/cooling of your apartment; and it is measured in BTU's, which can be directly converted to wattage, which is what you pay for. Since there are so many variables in doing an accurate heat-loss calculation, which were not included in your question, I gave you a simple "rule of thumb" which will bring you into the 'ballpark' but not tweak for an exact calculation. Yes, there are 'fixed' values- such as "R" value of insulation, energy rating of windows and doors, conductivity of surfaces(wood vs ceramic), which once you calculate, you can rule out.

However, the outside temperature, humidity, position and amount of sunlight, amount of heat sources(light, refrigeration), the effect of heating/cooling of other units which are all variables which have their relative effect on your "heat-loss" model, so much so that when you design a "heating/cooling" system, you adjust your capacity to meet whatever cooling demand is needed. Since you probably have a 'fixed' cfm system(variable speed fan motors are expensive and rare in residential unit) the tonnage unit itself is a ballpark range of the cooling needed.

The simple answer: Yes, you should always change your setpoint when the area is unoccupied. How much? It depends. Since most residential thermostats don't allow you to run an algorithim which changes your setpoint dependant on the variables I mentioned(an energy management system does, btw) you're stuck with a 'rule of thumb', which is what I gave you. And ultimately, you want the greatest cost savings, since you pay your utility and they offer a program of guarenteed savings, you've just short-cutted to the answer.

Even with humidity as a factor, you are going to recover by virtue of nightfall. and decrease of outdoor temperature. You should never run into a situation where you use more energy recovering than you save by turning up during unoccupied hours UNLESS, 1) your system is inefficient(condenser blocked/plugged/ dirty, filter dirty, inadequate charge(should be rare, unless you have a leak), 2) over use:(have a big party, lots of people, frequent in and out-in other words, more heat than was factored into heat/loss equation).

The thermostat is a 'crude' PID controller-although in cooling the anticipator is a fixed resistance, unlike heating. The "setpoint" is your "P", the anticipator(+/- 2F) is your "I", and the "D" is your room temperature, which is sensed and satisfied by the controller.

  • $\begingroup$ Thanks for the answer, I upvoted it. But it's still not what I am looking for. I am looking for a more mathematical model (with elements like the heat exchange (i.e. the ac), the heat sink (i.e. the apartment) and the environment). A general model which can be used for a first-order approximation. $\endgroup$ – Shashank Sawant Aug 31 '14 at 20:57

The main thing you want the air conditioner to do is to remove the heat that enters the apartment and eject it to outside.

The rate that the heat enters depends on the thermal conductivity of the apartment (which we assume to be fixed) and the temperature difference between the inside and the outside. The greater the heat difference, the greater the heat flux.

So during the time that no one is in the apartment, the closer to the outside temperature that the apartment is, the less heat enters. That is less heat that has to be ejected later. Any cooling that is done during the day will increase the temperature difference, increasing the heat flux, and increasing the work that has to be done.

I supposed there could be some loss due to PID overshoot, but I've never heard of a PID thermostat. Even the programmable ones tend to just modify the set point, not modify the behavior of the cooling system around the set point.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.