Why is it hot at the equator, cold at the poles?

Here’s a somewhat mathematical look at why it is hotter at the equator that at the poles. This is high school or basic college level science, using trigonometry (pre calculus), a slight step beyond the basic statement that the sun hits down more directly at the equator than at the poles. That’s the kid’s explanation, but we can understand better if we add a little math.

Solar radiation hits Detroit or any other non-equator point at an angle, As a result, less radiation power hits per square meter of land.

Lets use the diagram at right and trigonometry to compare the amount of sun-energy that falls on a square meter of land at the equator (0 latitude) and in a city at 42.5 N latitude (Detroit, Boston, and Rome are at this latitude). In each case, let’s consider high-noon on March 21 or September 20. These are the two equinox days, the only days each year when the day and night are equal length, and the only times when it is easy to calculate the angle of the sun as it deviates from the vertical by exactly the latitude on the days and times.

More specifically the equator is zero latitude, so on the equator at high-noon on the equinox, the sun will shine from directly overhead, or 0° from the vertical. Since the sun’s power in space is 1050 W/m2, every square meter of equator can expect to receive 1050 W of sun-energy, less the amount reflected off clouds and dust, or scattered off or air molecules (air scattering is what makes the sky blue). Further north, Detroit, Boston, and Rome sit at 42.5 latitude. At noon on March 31 the sun will strike earth at 42.5° from the vertical as shown in the lower figure above. From trigonometry, you can see that each square meter of these cities will receive cos 42.5 as much power as a square meter at the equator, except for any difference in clouds, dust, etc. Without clouds etc. that would be 1050 cos 42.5 = 774 W. Less sun power hits per square meter because each square meter is tilted. Earlier and later in the day, each spot will get less sunlight than at noon, but the proportion is the same, at least on one of the equinox days.

To calculate the likely temperature in Detroit, Boston, or Rome, I will use a simple energy balance. Ignoring heat storage in the earth for now, we will say that the heat in equals the heat out. We now ignore heat transfer by way of winds and rain, and approximate to say that the heat out leaves by black-body radiation alone, radiating into the extreme cold of space. This is not a very bad approximation since Black body radiation is the main temperature removal mechanism in most situations where large distances are involved. I’ve discussed black body radiation previously; the amount of energy radiated is proportional to luminosity, and to T4, where T is the temperature as measured in an absolute temperature scale, Kelvin or Rankin. Based on this, and assuming that the luminosity of the earth is the same in Detroit as at the equator,

T Detroit / Tequator  = √√ cos 42.5 = .927

I’ll now calculate the actual temperatures. For American convenience, I’ll choose to calculation in the Rankin Temperature scale, the absolute Fahrenheit scale. In this scale, 100°F = 560°R, 0°F = 460°R, and the temperature of space is 0°R as a good approximation. If the average temperature of the equator = 100°F = 38°C = 560°R, we calculate that the average temperature of Detroit, Boston, or Rome will be about .927 x 560 = 519°R = 59°F (15°C). This is not a bad prediction, given the assumptions. We can expect the temperature will be somewhat lower at night as there is no light, but the temperature will not fall to zero as there is retained heat from the day. The same reason, retained heat, explains why it is warmer will be warmer in these cities on September 20 than on March 31.

In the summer, these cities will be warmer because they are in the northern hemisphere, and the north pole is tilted 23°. At the height of summer (June 21) at high noon, the sun will shine on Detroit at an angle of 42.5 – 23° = 19.5° from the vertical. The difference in angle is why these cities are warmer on that day than on March 21. The equator will be cooler on that day (June 21) than on March 21 since the sun’s rays will strike the equator at 23° from the vertical on that day. These  temperature differences are behind the formation of tornadoes and hurricanes, with a tornado season in the US centering on May to July.

When looking at the poles, we find a curious problem in guessing what the average temperature will be. At noon on the equinox, the sun comes in horizontally, or at 90°from the vertical. We thus expect there is no warming power at all this day, and none for the six months of winter either. At first glance, you’d think the temperature at the poles would be zero, at least for six months of the year. It isn’t zero because there is retained heat from the summer, but still it makes for a more-difficult calculation.

To figure an average temperature of the poles, lets remember that during the 6 month summer the sun shines for 24 hours per day, and that the angle of the sun will be as high as 23° from the horizon, or 67° from the vertical for all 24 hours. Let’s assume that the retained heat from the summer is what keeps the temperature from falling too low in the winter and calculate the temperature at an .

Let’s assume that the sun comes in at the equivalent of 25° for the sun during the 6 month “day” of the polar summer. I don’t look at equinox, but rather the solar day, and note that the heating angle stays fixed through each 24 hour day during the summer, and does not decrease in the morning or as the afternoon wears on. Based on this angle, we expect that

TPole / Tequator  = √√ cos 65° = .806

TPole = .806 x 560°R = 452°R = -8°F (-22°C).

This, as it happens is 4° colder than the average temperature at the north pole, but not bad, given the assumptions. Maybe winds and water currents account for the difference. Of course there is a large temperature difference at the pole between the fall equinox and the spring equinox, but that’s to be expected. The average is, -4°F, about the temperature at night in Detroit in the winter.

One last thing, one that might be unexpected, is that temperature at the south pole is lower than at the north pole, on average -44°F. The main reason for this is that the snow on south pole is quite deep — more than 1 1/2 miles deep, with some rock underneath. As I showed elsewhere, we expect that, temperatures are lower at high altitude. Data collected from cores through the 1 1/2 mile deep snow suggest (to me) chaotic temperature change, with long ice ages, and brief (6000 year) periods of warm. The ice ages seem far worse than global warming.

Dr. Robert Buxbaum, December 30, 2017

Standing on the flat roof of my lab / factory building, I notice that virtually all of my neighbors’ roofs are black, covered by tar or bitumen. My roof was black too until three weeks ago; the roof was too hot to touch when I’d gone up to patch a leak. That’s not quite egg-frying hot, but I came to believe my repair would last longer if the roof stayed cooler. So, after sealing the leak with tar and bitumen, we added an aluminized over-layer from Ace hardware. The roof is cooler now than before, and I notice a major drop in air conditioner load and use.

My analysis of our roof coating follows; it’s for Detroit, but you can modify it for your location. Sunlight hits the earth carrying 1300 W/m2. Some 300W/m2 scatters as blue light (for why so much scatters, and why the sky is blue, see here). The rest, 1000 W/m2 or 308 Btu/ft2hr, comes through or reflects off clouds on a cloudy day and hits buildings at an angle determined by latitude, time of day, and season of the year.

Detroit is at 42° North latitude so my roof shows an angle of 42° to the sun at noon in mid spring. In summer, the angle is 20°, and in winter about 63°. The sun sinks lower on the horizon through the day, e.g. at two hours before or after noon in mid spring the angle is 51°. On a clear day, with a perfectly black roof, the heating is 308 Btu/ft2hr times the cosine of the angle.

To calculate our average roof heating, I integrated this heat over the full day’s angles using Euler’s method, and included the scatter from clouds plus an absorption factor for the blackness of the roof. The figure below shows the cloud cover for Detroit.

Average cloud cover for Detroit, month by month; the black line is the median cloud cover. On January 1, it is strongly overcast 60% of the time, and hardly ever clear; the median is about 98%. From http://weatherspark.com/averages/30042/Detroit-Michigan-United-States

Based on this and an assumed light absorption factor of σ = .9 for tar and σ = .2 after aluminum. I calculate an average of 105 Btu/ft2hr heating during the summer for the original black roof, and 23 Btu/ft2hr after aluminizing. Our roof is still warm, but it’s no longer hot. While most of the absorbed heat leaves the roof by black body radiation or convection, enough enters my lab through 6″ of insulation to cause me to use a lot of air conditioning. I calculate the heat entering this way from the roof temperature. In the summer, an aluminum coat is a clear winner.

High and Low Temperatures For Detroit, Month by Month. From http://weatherspark.com/averages/30042/Detroit-Michigan-United-States

Detroit has a cold winter too, and these are months where I’d benefit from solar heat. I find it’s so cloudy in winter that, even with a black roof, I got less than 5 Btu/ft2hr. Aluminizing reduced this heat to 1.2 Btu/ft2hr, but it also reduces the black-body radiation leaving at night. I should find that I use less heat in winter, but perhaps more in late spring and early fall. I won’t know the details till next year, but that’s the calculation.

The REB Research laboratory is located at 12851 Capital St., Oak Park, MI 48237. We specialize in hydrogen separations and membrane reactors. By Dr. Robert Buxbaum, June 16, 2013

What’s the quality of your home insulation

By Dr. Robert E. Buxbaum, June 3, 2013

It’s common to have companies call during dinner offering to blow extra insulation into the walls and attic of your home. Those who’ve added this insulation find a small decrease in their heating and cooling bills, but generally wonder if they got their money’s worth, or perhaps if they need yet-more insulation to get the full benefit. Here’s a simple approach to comparing your home heat bill to the ideal your home can reasonably reach.

The rate of heat transfer through a wall, Qw, is proportional to the temperature difference, ∆T, to the area, A, and to the average thermal conductivity of the wall, k; it is inversely proportional to the wall thickness, ∂;

Qw = ∆T A k /∂.

For home insulation, we re-write this as Qw = ∆T A/Rw where Rw is the thermal resistance of the wall, measured (in the US) as °F/BTU/hr-ft2. Rw = ∂/k.

Lets assume that your home’s outer wall thickness is nominally 6″ thick (0.5 foot). With the best available insulation, perfectly applied, the heat loss will be somewhat higher than if the space was filled with still air, k=.024 BTU/fthr°F, a result based on molecular dynamics. For a 6″ wall, the R value, will always be less than .5/.024 = 20.8 °F/BTU/hr-ft2.. It will be much less if there are holes or air infiltration, but for practical construction with joists and sills, an Rw value of 15 or 16 is probably about as good as you’ll get with 6″ walls.

To show you how to evaluate your home, I’ll now calculate the R value of my walls based on the size of my ranch-style home (in Michigan) and our heat bills. I’ll first do this in a simplified calculation, ignoring windows, and will then repeat the calculation including the windows. Windows are found to be very important. I strongly suggest window curtains to save heat and air conditioning,

The outer wall of my home is 190 feet long, and extends about 11 feet above ground to the roof. Multiplying these dimensions gives an outer wall area of 2090 ft2. I could now add the roof area, 1750 ft2 (it’s the same as the area of the house), but since the roof is more heavily insulated than the walls, I’ll estimate that it behaves like 1410 ft2 of normal wall. I calculate there are 3500 ftof effective above-ground area for heat loss. This is the area that companies keep offering to insulate.

Between December 2011 and February 2012, our home was about 72°F inside, and the outside temperature was about 28°F. Thus, the average temperature difference between the inside and outside was about 45°F; I estimate the rate of heat loss from the above-ground part of my house, Qu = 3500 * 45/R = 157,500/Rw.

Our house has a basement too, something that no one has yet offered to insulate. While the below-ground temperature gradient is smaller, it’s less-well insulated. Our basement walls are cinderblock covered with 2″ of styrofoam plus wall-board. Our basement floor is even less well insulated: it’s just cement poured on pea-gravel. I estimate the below-ground R value is no more than 1/2 of whatever the above ground value is; thus, for calculating QB, I’ll assume a resistance of Rw/2.

The below-ground area equals the square footage of our house, 1750 ft2 but the walls extend down only about 5 feet below ground. The basement walls are thus 950 ft2 in area (5 x 190 = 950). Adding the 1750 ft2 floor area, we find a total below-ground area of 2700 ft2.

The temperature difference between the basement and the wet dirt is only about 25°F in the winter. Assuming the thermal resistance is Rw/2, I estimate the rate of heat loss from the basement, QB = 2700*25*(2/Rw) = 135,000/Rw. It appears that nearly as much heat leaves through the basement as above ground!

Between December and February 2012, our home used an average of 597 cubic feet of gas per day or 25497 BTU/hour (heat value = 1025 BTU/ ft3). QU+ Q= 292,500/Rw. Ignoring windows, I estimate Rw of my home = 292,500/25497 = 11.47.

We now add the windows. Our house has 230 ft2 of windows, most covered by curtains and/or plastic. Because of the curtains and plastic, they would have an R value of 3 except that black-body radiation tends to be very significant. I estimate our windows have an R value of 1.5; the heat loss through the windows is thus QW= 230*45/1.5 = 6900 BTU/hr, about 27% of the total. The R value for our walls is now re-estimated to be 292,500/(25497-6900) = 15.7; this is about as good as I can expect given the fixed thickness of our walls and the fact that I can not easily get an insulation conductivity lower than still air. I thus find that there will be little or no benefit to adding more above-ground wall insulation to my house.

To save heat energy, I might want to coat our windows in partially reflective plastic or draw the curtains to follow the sun. Also, since nearly half the heat left from the basement, I may want to lay a thicker carpet, or lay a reflective under-layer (a space blanket) beneath the carpet.

To improve on the above estimate, I could consider our furnace efficiency; it is perhaps only 85-90% efficient, with still-warm air leaving up the chimney. There is also some heat lost through the door being opened, and through hot water being poured down the drain. As a first guess, these heat losses are balanced by the heat added by electric usage, by the body-heat of people in the house, and by solar radiation that entered through the windows (not much for Michigan in winter). I still see no reason to add more above-ground insulation. Now that I’ve analyzed my home, it’s time for you to analyze yours.

Chaos, Stocks, and Global Warming

Two weeks ago, I discussed black-body radiation and showed how you calculate the rate of radiative heat transfer from any object. Based on this, I claimed that basal metabolism (the rate of calorie burning for people at rest) was really proportional to surface area, not weight as in most charts. I also claimed that it should be near-impossible to lose weight through exercise, and went on to explain why we cover the hot parts of our hydrogen purifiers and hydrogen generators in aluminum foil.

I’d previously discussed chaos and posted a chart of the earth’s temperature over the last 600,000 years. I’d now like to combine these discussions to give some personal (R. E. Buxbaum) thoughts on global warming.

Black-body radiation differs from normal heat transfer in that the rate is proportional to emissivity and is very sensitive to temperature. We can expect the rate of heat transfer from the sun to earth will follow these rules, and that the rate from the earth will behave similarly.

That the earth is getting warmer is seen as proof that the carbon dioxide we produce is considered proof that we are changing the earth’s emissivity so that we absorb more of the sun’s radiation while emitting less (relatively), but things are not so simple. Carbon dioxide should, indeed promote terrestrial heating, but a hotter earth should have more clouds and these clouds should reflect solar radiation, while allowing the earth’s heat to radiate into space. Also, this model would suggest slow, gradual heating beginning, perhaps in 1850, but the earth’s climate is chaotic with a fractal temperature rise that has been going on for the last 15,000 years (see figure).

Recent temperature variation as measured from the Greenland Ice. Like the stock market, it shows aspects of chaos.

Over a larger time scale, the earth’s temperature looks, chaotic and cyclical (see the graph of global temperature in this post) with ice ages every 120,000 years, and chaotic, fractal variation at times spans of 100 -1000 years. The earth’s temperature is self-similar too; that is, its variation looks the same if one scales time and temperature. This is something that is seen whenever a system possess feedback and complexity. It’s seen also in the economy (below), a system with complexity and feedback.

Manufacturing Profit is typically chaotic — and seems to have cold spells very similar to the ice ages seen above.

The economy of any city is complex, and the world economy even more so. No one part changes independent of the others, and as a result we can expect to see chaotic, self-similar stock and commodity prices for the foreseeable future. As with global temperature, the economic data over a 10 year scale looks like economic data over a 100 year scale. Surprisingly,  the economic data looks similar to the earth temperature data over a 100 year or 1000 year scale. It takes a strange person to guess either consistently as both are chaotic and fractal.

It takes a rather chaotic person to really enjoy stock trading (Seen here, Gomez Addams of the Addams Family TV show).

Clouds and ice play roles in the earth’s feedback mechanisms. Clouds tend to increase when more of the sun’s light heats the oceans, but the more clouds, the less heat gets through to the oceans. Thus clouds tend to stabilize our temperature. The effect of ice is to destabilize: the more heat that gets to the ice, the more melts and the less of the suns heat is reflected to space. There is time-delay too, caused by the melting flow of ice and ocean currents as driven by temperature differences among the ocean layers, and (it seems) by salinity. The net result, instability and chaos.

The sun has chaotic weather too. The rate of the solar reactions that heat the earth increases with temperature and density in the sun’s interior: when a volume of the sun gets hotter, the reaction rates pick up making the volume yet-hotter. The temperature keeps rising, and the heat radiated to the earth keeps increasing, until a density current develops in the sun. The hot area is then cooled by moving to the surface and the rate of solar output decreases. It is quite likely that some part of our global temperature rise derives from this chaotic variation in solar output. The ice caps of Mars are receding.

The change in martian ice could be from the sun, or it might be from Martian dust in the air. If so, it suggests yet another feedback system for the earth. When economic times age good we have more money to spend on agriculture and air pollution control. For all we know, the main feedback loops involve dust and smog in the air. Perhaps, the earth is getting warmer because we’ve got no reflective cloud of dust as in the dust-bowl days, and our cities are no longer covered by a layer of thick, black (reflective) smog. If so, we should be happy to have the extra warmth.

Most Heat Loss Is Black-Body Radiation

In a previous post I used statistical mechanics to show how you’d calculate the thermal conductivity of any gas and showed why the insulating power of the best normal insulating materials was usually identical to ambient air. That analysis only considered the motion of molecules and not of photons (black-body radiation) and thus under-predicted heat transfer in most circumstances. Though black body radiation is often ignored in chemical engineering calculations, it is often the major heat transfer mechanism, even at modest temperatures.

One can show from quantum mechanics that the radiative heat transfer between two surfaces of temperature T and To is proportional to the difference of the fourth power of the two temperatures in absolute (Kelvin) scale.

Here Pnet is the net heat transfer rate, A is the area of the surfaces, σ is the Stefan–Boltzmann constantε is the surface emissivity, a number that is 1 for most non-metals and .3 for stainless steel.  For A measured in m2σ = 5.67×10−8 W m−2 K−4.

Unlike with conduction, heat transfer does not depend on the distances between the surfaces but only on the temperature and the infra-red (IR) reflectivity. This is different from normal reflectivity as seen in the below infra-red photo of a lightly dressed person standing in a normal room. The fellow has a black plastic bag on his arm, but you can hardly see it here, as it hardly affects heat loss. His clothes, don’t do much either, but his hair and eyeglasses are reasonably effective blocks to radiative heat loss.

 Infrared picture of a fellow wearing a black plastic bag on his arm. The bag is nearly transparent to heat radiation, while his eyeglasses are opaque. His hair provides some insulation.

As an illustrative example, lets calculate the radiative and conductive heat transfer heat transfer rates of the person in the picture, assuming he has  2 m2 of surface area, an emissivity of 1, and a body and clothes temperature of about 86°F; that is, his skin/clothes temperature is 30°C or 303K in absolute. If this person stands in a room at 71.6°F, 295K, the radiative heat loss is calculated from the equation above: 2 *1* 5.67×10−8 * (8.43×109 -7.57×109) = 97.5 W. This is 23.36 cal/second or 84.1 Cal/hr or 2020 Cal/day; this is nearly the expected basal calorie use of a person this size.

The conductive heat loss is typically much smaller. As discussed previously in my analysis of curtains, the rate is inversely proportional to the heat transfer distance and proportional to the temperature difference. For the fellow in the picture, assuming he’s standing in relatively stagnant air, the heat boundary layer thickness will be about 2 cm (0.02m). Multiplying the thermal conductivity of air, 0.024 W/mK, by the surface area and the temperature difference and dividing by the boundary layer thickness, we find a Wattage of heat loss of 2*.024*(30-22)/.02 = 19.2 W. This is 16.56 Cal/hr, or 397 Cal/day: about 20% of the radiative heat loss, suggesting that some 5/6 of a sedentary person’s heat transfer may be from black body radiation.

We can expect that black-body radiation dominates conduction when looking at heat-shedding losses from hot chemical equipment because this equipment is typically much warmer than a human body. We’ve found, with our hydrogen purifiers for example, that it is critically important to choose a thermal insulation that is opaque or reflective to black body radiation. We use an infra-red opaque ceramic wrapped with aluminum foil to provide more insulation to a hot pipe than many inches of ceramic could. Aluminum has a far lower emissivity than the nonreflective surfaces of ceramic, and gold has an even lower emissivity at most temperatures.

Many popular insulation materials are not black-body opaque, and most hot surfaces are not reflectively coated. Because of this, you can find that the heat loss rate goes up as you add too much insulation. After a point, the extra insulation increases the surface area for radiation while barely reducing the surface temperature; it starts to act like a heat fin. While the space-shuttle tiles are fairly mediocre in terms of conduction, they are excellent in terms of black-body radiation.

There are applications where you want to increase heat transfer without having to resort to direct contact with corrosive chemicals or heat-transfer fluids. Often black body radiation can be used. As an example, heat transfers quite well from a cartridge heater or band heater to a piece of equipment even if they do not fit particularly tightly, especially if the outer surfaces are coated with black oxide. Black body radiation works well with stainless steel and most liquids, but most gases are nearly transparent to black body radiation. For heat transfer to most gases, it’s usually necessary to make use of turbulence or better yet, chaos.