Facing broken central air, extreme heat, and requirements to move away from freon, I was wanting to know if Air conditioner BTU ratings are even accurate and why they are so hard to size right. I uncovered a 2009 DOE lab test report* that put 2 mini-split AC units through testing to verify their claimed specifications. Both units were of Japanese manufacture.
I whittled down the report to focus on what matters to me, which is simply the cooling power, or BTUs. SEER ratings are also important, to save electricity, but what is the point of efficiency if there isn’t enough cooling power? Manufacturers state cooling capacity, which they will quote as BTUs (British Thermal Units), for example, 10,000 BTUs capacity. But the actual units are BTUs/hour. Why they feel they should leave out their denominator I’ll never appreciate.
At any rate, the units in the DOE paper give a manufacturer’s stated range such as 2,800 to 12,000 BTU/h. As a consumer, I have never seen a range. They show that this ranges from the Lo to Med and Hi fan setting, so I guess you better set your AC on Hi fan, because I don’t believe Lo saves electricity proportionally, but I could be wrong about that.
Whatever the meaning of this super wide range may be, I need a quick and dirty, easy understanding for my (mind’s) comfort. Page 17 of the report refers to “Cooling test results…Steady State”. That’s where I focused. They say “As expected, the cooling capacity decreases with increasing outdoor temperature “.
The study shows in essence that the cooling capacity of the units are at least what they were claimed to be. I’m happy that this study reveals that per true physics definitions and by independent study that in fact at least some ACs do what they say. Regarding BTUs, I gather that based on the quote above, that AC units remove less BTUs from a volume of air if the temperature differential between inside and outside is greater. My guess is that this interesting fact is because the refrigerant line warms up more in this case.
As a final determination, I would like to get a feel for how many BTUs need to be removed to make my otherwise 85-degree F room comfortable. I want to avoid what to me is a befuddling mish mosh of sizing info provided by suppliers and installers that I have found to be absolutely useless in the past.
Therefore, using the definition of a BTU**
the quantity of heat required to raise the temperature of one pound of liquid water by 1 degree Fahrenheit at the temperature that water has its greatest density (approximately 39 degrees Fahrenheit)
I propose to estimate how much water is in air and build a model based on that.
The estimate I use is “the quantity of water vapor per cubic foot is 0.00395 lb, if the air is … 50% saturated” .*** Therefore, if I have a room that is 25’x30’x8’ (6000 cubic feet), it typically holds about 24 pounds of water in the air. If that water were at 85 degrees F and I want to get it to 75 F in say, 2 hours, I would need to remove about 24 * (10/2) *6000 = 720,000 BTUs! A crazy amount of BTUs!
Knowing that air isn’t water, it’s gas with a little water in it, what I need is a fudge factor! Working backwards, I guesstimate based on my personal experience with window units is that an appropriate answer for this room would be 15,000 -20,000 BTUs (per hour). The factor of adjustment (fudge) then would be the inverse of 720,000/15,000, or 1/48. I therefore will go ahead and multiply the cube footage CUBEFT of a room I have needing cooling by
[(CUBEFT *0.00395) * CUBEFT *5]/48
= 0.00042 * CUBEFT ^ 2
and believe that is on the order of what is needed to get one room chilled by one mini-split in 2 hours.
The ‘cool’ significance (intended) of this admittedly oversimplified estimate is that the result is a parabolic (squared) function of cubic-feet. And if I’m wrong, drink lots of tea.
-Robert
* https://www.nrel.gov/docs/fy11osti/52175.pdf
** eia.gov
