2
$\begingroup$

I'm need to understand the following: to keep the room at confortable temperature (70 degree, for example), how does the amount of energy consumed by the AC grow as the outside air temperature rises in the summer over the same length of time (a day, say)? Is it more or less linear? quadratic? exponential? or anything else?

I posted the same question on another StackExchange site, but I figured that physicist may provide more help because it seems to me that it is related to both heat conduction and the working mechanism of AC.

Thanks.

$\endgroup$
1

2 Answers 2

1
$\begingroup$

Without getting too complicated I'd say it's linear.

Two bodies in contact exchange heat at a rate that's proportional to the temperature difference between them. The rate of heat transfer in with an outside temperature of 80 degrees will be twice the rate at 75 degrees (assuming your 70-degree room temperature).

Your AC needs to pump this heat out as fast as it comes in to maintain 70 degrees, so it will need to do it twice as fast when it's 80 degrees outsite than when it's 75 degrees.

This is a linear relationship.

$\endgroup$
8
  • 2
    $\begingroup$ An ideal AC is a Carnot machine running backward. So it does not work really linearly. Some practical things like friction make it definitely nonlinear. $\endgroup$
    – Georg
    Apr 13, 2011 at 22:20
  • $\begingroup$ @Georg: Good point. But isn't the efficiency difference only about 5% over the range of temperatures we're talking about? $\endgroup$
    – John
    Apr 13, 2011 at 22:36
  • 3
    $\begingroup$ If the AC had perfect carnot efficiency, it would be quadratic, i.e. the energy cost to move a unit of heat against a temperature increase, is proportional to that increase. Because of friction and heat loses etc. it is most likely close to linear at least for small differences in temperature. $\endgroup$ Apr 13, 2011 at 22:38
  • $\begingroup$ Also the heat load coming in will be strongly affected by things other than the outside temperature. Shortwave radiation coming through windows. Shortwave (sunlight) and longwave radiation heating outside walls possibly to several tens of degrees above the ambient outdoor temperature. $\endgroup$ Apr 13, 2011 at 22:40
  • 1
    $\begingroup$ Jerry, I was making the assumption that deltaT is much smaller than the absolute temperature. I.E. throw out the next higher order term in the expansion. $\endgroup$ Apr 14, 2011 at 4:19
0
$\begingroup$

The answer will vary significantly with the construction of the home or room in question. The amount and rating of insulation, AC system type and efficiency, the volume of external air that is exchanged with the inside (which is substantial), windspeed, angle of the sun, potential shade, the existence of internal heat sinks (such as a large basement) must all be considered. Every variable could drastically change the slope and form of the overall function of heat pump input energy as it relates to exterior air temperature.

The percentage of surface area exposed to the external heat is likely the most significant variable. A perfect 6-sided cube of a home, for instance, would have 5/6th of its surface area exposed to heat. The bottom of the cube, as it's against the ground with a different temperature, may act as a sink. On other hand, a squashed rambler may have a much larger roof area that could, in extreme cases, mean just under 1/2 of the total surface area is exposed to the heat, in relation to the potential basement/floor sink.

The ability of the AC condenser unit to release its absorbed heat is a massive factor contributing to energy use per BTU. A clogged fan or debris around the condenser could ruin efficiency, especially at high duty cycle.

Long story short, you cannot take such a tremendously dynamic system and simplify it in a way that would reasonably apply to an imaginary situation. The only basic concepts you could rely upon are conservation of energy and Newton's 2nd law.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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