A Lesson in Average Temperature Calculations

June 28, 2013

28 Jun (PRINCIPIA SCIENTIFIC)Say your house has two floors.  Downstairs the temperature is at 72°, upstairs at 76°. You might conclude, then, that the house’s average interior temperature is 74. But wait. Now you recall that the upstairs is 15% smaller. So should the average temperature be estimated thus?

Introduction: The “Two Floors” problem

by Alan Siddons

Whole house = 100%
Downstairs = X
Upstairs = X × 0.85
Therefore X + 0.85 X = 100, meaning that
1.85 X = 100
So X = 54.054

Thus

Downstairs = 54.054% of the house
Upstairs = 45.946% of the house
“Weighting” the two temperatures, then…
Downstairs = 72 × 0.54054 = 38.91888
Upstairs = 76 × 0.45946 = 34.91896

Adding these two numbers, the house’s actual average is therefore closer to 73.84°. Volume or area must always be factored in.

See how complicated an “average temperature” can be? And you haven’t even counted the crawl space in the attic! Finding an average temperature is more difficult with the Stefan-Boltzmann equation [1], however, because there’s a 4th root involved.

Complication #1: The 4th root problem

Say that two spots on your blackbody sphere are being exposed to 50 and 100 watts per square meter. (Due to curvature, remember, a single light source gets spread out and becomes weaker.) Using the Stefan-Boltzmann equation the two temperatures will be about 172 and 205 Kelvin respectively, i.e., an average of 188.5 K. But the average irradiance is 75 W/m², which corresponds to 191 K. That’s 2.5 degrees off the mark. In other words, average temperature does not agree with average irradiance, and vice versa.

Take three spots at 100, 200, and 300 W/m². The average of course is 200 W/m². The temperatures are 205, 244, and 270 respectively, averaging about 240 K. But 200 W/m², the average, equals 244 K. Now you’re 4 degrees off the mark. And so on, as you proceed to compare irradiance with temperature on each and every angle of a half-lit sphere. It’s a huge problem to tackle. Throw in rotation (i.e., the irradiance is constantly changing) and the heat-retention of various three-dimensional substances, and the problem runs out of control.

Complication #2: The minus 18 problem

As for the famous minus 18° C surface temperature that earth is supposed to have without the greenhouse effect, that figure assumes a blackbody surface absorbing about 239 W/m² “on average.” But check the Kiehl-Trenberth chart [2]. Due to clouds and other obscuring factors, the actual surface average is given as only 168 W/m². That figure corresponds to minus 40° C on the surface, meaning that it has to rise by 55 degrees, not 33, in order to reach the accepted average of plus 15. Anyone who tells you, then, that the ‘greenhouse effect’ makes the earth’s surface 33 degrees warmer is merely confessing his (or her) own ignorance.

Conclusion: The average temperature without temperatures problem

Ask yourself what kind of “average temperature” consists of no highs and lows and in-betweens? The earth’s purported average temperature from Stefan-Boltzmann lacks any specifics, no information about average polar vs. equatorial differences — no information even about average day and night differences. What sort of average is that? This is why NASA engineers couldn’t find any use for it [3]. And as I say, it’s because it really isn’t an average temperature in the first place, it’s merely the result of dividing irradiance by 4 and thoughtlessly parroting what an equation says.

References:

[1] http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html

[2] http://www.cgd.ucar.edu/cas/abstracts/files/kevin1997_1.html

[3] http://www.tech-know-group.com/papers/Greenhouse_Effect_on_the_Moon.pdf

Tags: , , , , , , , , , ,

You must be logged in to comment

Log in