Tag Archives: heat

Heat conduction in insulating blankets, aerogels, space shuttle tiles, etc.

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A lot about heat conduction in insulating blankets can be explained by the ordinary motion of gas molecules. That’s because the thermal conductivity of air (or any likely gas) is much lower than that of glass, alumina, or any likely solid material used for the structure of the blanket. At any temperature, the average kinetic energy of an air molecule is 1/2kT in any direction, or 3/2kT altogether; where k is Boltzman’s constant, and T is absolute temperature, °K. Since kinetic energy equals 1/2 mv2, you find that the average velocity in the x direction must be v = √kT/m = √RT/M. Here m is the mass of the gas molecule in kg, M is the molecular weight also in kg (0.029 kg/mol for air), R is the gas constant 8.29J/mol°C, and v is the molecular velocity in the x direction, in meters/sec. From this equation, you will find that v is quite large under normal circumstances, about 290 m/s (650 mph) for air molecules at ordinary temperatures of 22°C or 295 K. That is, air molecules travel in any fixed direction at roughly the speed of sound, Mach 1 (the average speed including all directions is about √3 as fast, or about 1130 mph).

The distance a molecule will go before hitting another one is a function of the cross-sectional areas of the molecules and their densities in space. Dividing the volume of a mol of gas, 0.0224 m3/mol at “normal conditions” by the number of molecules in the mol (6.02 x10^23) gives an effective volume per molecule at this normal condition: .0224 m3/6.0210^23 = 3.72 x10^-26 m3/molecule at normal temperatures and pressures. Dividing this volume by the molecular cross-section area for collisions (about 1.6 x 10^-19 m2 for air based on an effective diameter of 4.5 Angstroms) gives a free-motion distance of about 0.23×10^-6 m or 0.23µ for air molecules at standard conditions. This distance is small, to be sure, but it is 1000 times the molecular diameter, more or less, and as a result air behaves nearly as an “ideal gas”, one composed of point masses under normal conditions (and most conditions you run into). The distance the molecule travels to or from a given surface will be smaller, 1/√3 of this on average, or about 1.35×10^-7m. This distance will be important when we come to estimate heat transfer rates at the end of this post.

 

Molecular motion of an air molecule (oxygen or nitrogen) as part of heat transfer process; this shows how some of the dimensions work.

Molecular motion of an air molecule (oxygen or nitrogen) as part of heat transfer process; this shows how some of the dimensions work.

The number of molecules hitting per square meter per second is most easily calculated from the transfer of momentum. The pressure at the surface equals the rate of change of momentum of the molecules bouncing off. At atmospheric pressure 103,000 Pa = 103,000 Newtons/m2, the number of molecules bouncing off per second is half this pressure divided by the mass of each molecule times the velocity in the surface direction. The contact rate is thus found to be (1/2) x 103,000 Pa x 6.02^23 molecule/mol /(290 m/s. x .029 kg/mol) = 36,900 x 10^23 molecules/m2sec.

The thermal conductivity is merely this number times the heat capacity transfer per molecule times the distance of the transfer. I will now calculate the heat capacity per molecule from statistical mechanics because I’m used to doing things this way; other people might look up the heat capacity per mol and divide by 6.02 x10^23: For any gas, the heat capacity that derives from kinetic energy is k/2 per molecule in each direction, as mentioned above. Combining the three directions, that’s 3k/2. Air molecules look like dumbbells, though, so they have two rotations that contribute another k/2 of heat capacity each, and they have a vibration that contributes k. We begin with an approximate value for k = 2 cal/mol of molecules per °C; it’s actually 1.987 but I round up to include some electronic effects. Based on this, we calculate the heat capacity of air to be 7 cal/mol°C at constant volume or 1.16 x10^-23 cal/molecule°C. The amount of energy that can transfer to the hot (or cold) wall is this heat capacity times the temperature difference that molecules carry between the wall and their first collision with other gases. The temperature difference carried by air molecules at standard conditions is only 1.35 x10-7 times the temperature difference per meter because the molecules only go that far before colliding with another molecule (remember, I said this number would be important). The thermal conductivity for stagnant air per meter is thus calculated by multiplying the number of molecules times that hit per m2 per second, the distance the molecule travels in meters, and the effective heat capacity per molecule. This would be 36,900 x 10^23  molecules/m2sec x 1.35 x10-7m x 1.16 x10^-23 cal/molecule°C = 0.00578 cal/ms°C or .0241 W/m°C. This value is (pretty exactly) the thermal conductivity of dry air that you find by experiment.

I did all that math, though I already knew the thermal conductivity of air from experiment for a few reasons: to show off the sort of stuff you can do with simple statistical mechanics; to build up skills in case I ever need to know the thermal conductivity of deuterium or iodine gas, or mixtures; and finally, to be able to understand the effects of pressure, temperature and (mainly insulator) geometry — something I might need to design a piece of equipment with, for example, lower thermal heat losses. I find, from my calculation that we should not expect much change in thermal conductivity with gas pressure at near normal conditions; to first order, changes in pressure will change the distance the molecule travels to exactly the same extent that it changes the number of molecules that hit the surface per second. At very low pressures or very small distances, lower pressures will translate to lower conductivity, but for normal-ish pressures and geometries, changes in gas pressure should not affect thermal conductivity — and does not.

I’d predict that temperature would have a larger effect on thermal conductivity, but still not an order-of magnitude large effect. Increasing the temperature increases the distance between collisions in proportion to the absolute temperature, but decreases the number of collisions by the square-root of T since the molecules move faster at high temperature. As a result, increasing T has a √T positive effect on thermal conductivity.

Because neither temperature nor pressure has much effect, you might expect that the thermal conductivity of all air-filed insulating blankets at all normal-ish conditions is more-or-less that of standing air (air without circulation). That is what you find, for the most part; the same 0.024 W/m°C thermal conductivity with standing air, with high-tech, NASA fiber blankets on the space shuttle and with the cheapest styrofoam cups. Wool felt has a thermal conductivity of 0.042 W/m°C, about twice that of air, a not-surprising result given that wool felt is about 1/2 wool and 1/2 air.

Now we can start to understand the most recent class of insulating blankets, those with very fine fibers, or thin layers of fiber (or aluminum or gold). When these are separated by less than 0.2µ you finally decrease the thermal conductivity at room temperature below that for air. These layers decrease the distance traveled between gas collisions, but still leave the same number of collisions with the hot or cold wall; as a result, the smaller the gap below .2µ the lower the thermal conductivity. This happens in aerogels and some space blankets that have very small silica fibers, less than .1µ apart (<100 nm). Aerogels can have much lower thermal conductivities than 0.024 W/m°C, even when filled with air at standard conditions.

In outer space you get lower thermal conductivity without high-tech aerogels because the free path is very long. At these pressures virtually every molecule hits a fiber before it hits another molecule; for even a rough blanket with distant fibers, the fibers bleak up the path of the molecules significantly. Thus, the fibers of the space shuttle (about 10 µ apart) provide far lower thermal conductivity in outer space than on earth. You can get the same benefit in the lab if you put a high vacuum of say 10^-7 atm between glass walls that are 9 mm apart. Without the walls, the air molecules could travel 1.3 µ/10^-7 = 13m before colliding with each other. Since the walls of a typical Dewar are about 0.009 m apart (9 mm) the heat conduction of the Dewar is thus 1/150 (0.7%) as high as for a normal air layer 9mm thick; there is no thermal conductivity of Dewar flasks and vacuum bottles as such, since the amount of heat conducted is independent of gap-distance. Pretty spiffy. I use this knowledge to help with the thermal insulation of some of our hydrogen generators and hydrogen purifiers.

There is another effect that I should mention: black body heat transfer. In many cases black body radiation dominates: it is the reason the tiles are white (or black) and not clear; it is the reason Dewar flasks are mirrored (a mirrored surface provides less black body heat transfer). This post is already too long to do black body radiation justice here, but treat it in more detail in another post.

RE. Buxbaum

Small hydrogen generators for cooling dynamo generators

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A majority of the electricity used in the US comes from rotating dynamos. Power is provided to the dynamos by a turbine or IC engine and the dynamo turns this power into electricity by moving a rotating coil (a rotor) through a non-rotating magnetic field provided by magnets or a non-rotating coil (a stator). While it is easy to cool the magnets or stator, cooling the rotor is challenging as there is no possibility to connect it cooling water or heat transfer paste. One of the more common options is hydrogen gas.

It is common to fill the space between the rotor and the stator with hydrogen gas. Heat transfers from the rotor to the stator or to the walls of the dynamo through the circulating hydrogen. Hydrogen has the lowest density of any gas, and the highest thermal conductivity of any gas. The low density is important because it reduces the power drag (wind drag) on the rotor. The high heat transfer coefficient helps cool the rotor so that it does not burn out at high power draw.

Hydrogen is typically provided to the dynamo by a small hydrogen generator or hydrogen bottle. While we have never sold a hydrogen generator to this market, I strongly believe that our membrane reactor hydrogen generators would be competitive; the cost of hydrogen is lower than that of bottled gas; it is far more convenient and safe; and the hydrogen is purer than from electrolysis.

Why isn’t the sky green?

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Yesterday I blogged with a simple version of why the sky was blue and not green. Now I’d like to add mathematics to the treatment. The simple version said that the sky was blue because the sun color was a spectrum centered on yellow. I said that molecules of air scattered mostly the short wavelength, high frequency light colors, indigo and blue. This made the sky blue. I said that, the rest of the sunlight was not scattered, so that the sun looked yellow. I then said that the only way for the sky to be green would be if the sun were cooler, orange say, then the sky would be green. The answer is sort-of true, but only in a hand-waving way; so here’s the better treatment.

Light scatters off of dispersed small particles in proportion to wavelength to the inverse 4th power of the wavelength. That is to say, we expect air molecules will scatter more short wavelength, cool colors (purple and indigo) than warm colors (red and orange) but a real analysis must use the actual spectrum of sunlight, the light power (mW/m2.nm) at each wavelength.

intensity of sunlight as a function of wavelength (frequency)

intensity of sunlight as a function of wavelength

The first thing you’ll notice is that the light from our sun isn’t quite yellow, but is mostly green. Clearly plants understand this, otherwise chlorophyl would be yellow. There are fairly large components of blue and red too, but my first correction to the previous treatment is that the yellow color we see as the sun is a trick of the eye called additive color. Our eyes combine the green and red of the sun’s light, and sees it as yellow. There are some nice classroom experiment you can do to show this, the simplest being to make a Maxwell top with green and red sections, spin the top, and notice that you see the color as yellow.

In order to add some math to the analysis of sky color, I show a table below where I divided the solar spectrum into the 7 representative colors with their effective power. There is some subjectivity to this, but I took red as the wavelengths from 620 to 750nm so I claim on the table was 680 nm. The average power of the red was 500 mW/m2nm, so I calculate the power as .5 W/m2nm x 130 nm = 65W/m2. Similarly, I took orange to be the 30W/m2 centered on 640nm, etc. This division is presented in the first 3 columns of the following table. The first line of the table is an approximate of the Rayleigh-scatter factor for our atmosphere, with scatter presented as the percent of the incident light. That is % scattered = 9E11/wavelength^4.skyblue scatter

To use the Rayleigh factor, I calculate the 1/wavelength of each color to the 4th power; this is shown in the 4th column. The scatter % is now calculated and I apply this percent to the light intensities to calculate the amount of each color that I’d expect in the scattered and un-scattered light (the last two columns). Based on this, I find that the predominant wavelength in the color of the sky should be blue-cyan with significant components of green, indigo, and violet. When viewed through a spectroscope, I find that these are the colors I see (I have a pocket spectroscope and used it an hour ago to check). Viewed through the same spectroscope (with eye protection), I expect the sun should look like a combination of green and red, something our eyes see as yellow (I have not done this personally). At any rate, it appears that the sky looks blue because our eyes see the green+ cyan+ indigo + purple in the scattered light as sky blue.220px-RGB_illumination

At sunrise and sunset when the sun is on the horizon the scatter percents will be higher, so that all of the sun’s colors will be scattered except red and orange. The sun looks orange then, as expected, but the sky should look blue-green, as that’s the combination of all the other colors of sunlight when orange and red are removed. I’ve not checked this last yet. I’ll have to take my spectroscope to a fine sunset and see what I see when I look at the sky.

Why isn’t the sky green and the sun orange?

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Part of the reason the sky isn’t green has to do with the color of the sun. The sun’s color, and to a lesser extent, the sky color both are determined by the sun’s surface color, yellow. This surface color results from black body radiation: if you heat up a black object it will first glow red, then orange, yellow, green etc. Red is a relatively cool color because it’s a low frequency (long wavelength) and low frequencies are the lowest energy photons, and thus are the easiest for a black body to produce. As one increases the temperature of a black object, the total number of photons increases for all wavelengths, but the short wavelength (high frequency) colors increase faster than the of long wavelength colors. As a result, the object is seen to change color to orange, then yellow, or to any other color representative of objects at that particular temperature.

Our star is called a yellow sun because the center color of its radiation is yellow. The sun provides radiation in all colors and wavelengths, even colors invisible to the eye, infra red and ultra violet, but because of its temperature, most of the radiated energy appears as yellow. This being said, you may wonder why the sky isn’t yellow (the sky of Mars mostly is).

The reason the sky is blue, is that some small fraction of the light of the sun (about 10%) scatters off of the molecules of the air. This is called Rayleigh scatter — the scatter of large wavelegth waves off of small objects.  The math for this will be discussed in another post, but the most relevant aspect here is that the fraction that is scattered is proportional to the 4th power of the frequency. This is to say, that the high frequencies (blue, indigo, and violet) scatter a lot, about 20%, while the red hardly scatters at all. As a result the sky has a higher frequency color than the sun does. In our case, the sky looks blue, while the sun looks slightly redder from earth than it does from space — at least that’s the case for most of the day.

The sun looks orange-red at sundown because the sunlight has to go through more air. Because of this, a lot more of the yellow, green, and blue scatter away before we see it. Much more of the scatter goes off into space, with the result that the sky to looks dark, and somewhat more greenish at sundown. If the molecules were somewhat bigger, we’d still see a blue sky, maybe somewhat greener, with a lot more intensity. That’s the effect that carbon dioxide has — it causes more sunlight to scatter, making the sky brighter. If the sun were cooler (orange say), the sky would appear green. That’s because there would be less violet and blue in the sunlight, and the sky color would be shifted to the longer wavelengths. On planets where the sun is cooler than ours, the sky is likely green, but could be yellow or red.

Rayleigh scatter requires objects that are much smaller than the light wavelength. A typical molecule of air is about 1 nm in size (1E-9 of a meter), while the wavelength of yellow light is 580 nm. That’s much larger than the size of air molecules. Snow appears white because the size of the crystals are the size of the sun wavelengths, tor bigger, 500-2000 nm. Thus, the snow looks like all the colors of the sun together, and that’s white. White = the sum of all the colors: red + orange + blue + green + yellow + violet + indigo.

Robert Buxbaum  Jan. 27, 2013 (revised)

The joy of curtains

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By Dr. Robert E. Buxbaum January 18, 2013

In our northern climates most homes have double-paned windows; they cost a fortune, and are a lot better than plain glass, but they still lose a lot of heat: far more than the equivalent area of wall. The insulation value is poor mostly because the thickness is low: a typical double pane window is only ½” thick. The glass panes have hardly any insulation value, so the majority of the insulation is the 0.3″ air space between them. Our outer walls, by contrast, are typically 6” thick filled with glass –wool. The wall is 12 times as thick as the window, and it turns out that the R value is about 12 times as great. Since window area is about 1/10 the wall area, we can expect that about half your homes heat goes out through the windows (about half the air-conditioner cooling in the summer too). A good trick to improve your home’s insulation, then, is to add curtains as this provides a fairly thick layer of stagnant air inside the room, right next to your windows.

To see how much you can save by adding curtains, it’s nice (for me, and my mind-set mostly) to talk in terms of R values. In the northern USA, the “R” value of a typical, well-insulated outer wall is about 24. What that means is that it takes 24°F and one square foot of wall to remove 1 BTU per hour. That is, the resistance to heat loss is 24 °F.hr.ft2/BTU. The R value for a typical double pane window is about 2 in the same units, and is only 1 if you have single panes. The insulating quality of our windows is so poor that, for many homes, more heat is lost through the windows than through the rest of the wall space.

To figure out how much heat is lost through your windows take the area in square feet multiply by a typical temperature differential (50°F might be typical in Michigan), and divide by the R value of your paned windows (1 or 2) depending on whether it’s single or double paned. Since heat costs about $10/MMBTU ($10 per million BTU) for a gas heated house, you can figure out what a small, 10 ft2 window costs a typical Michigan householder as follows, assuming a single pane (R=1):

Q = Area* ∆T/R = 10 ft2 * 50°F/1 = 500 BTU/hr. Here Q is the heat lost per unit time, ∆T is the temperature difference between the window surface and the room, and A is the ara of the window surface.

Since there are 24 hours in a day, and 30.5 days in a month the dollar cost of that window is 500*24*31.5*10/1,000,000 = $3.78/month. After a few years, you’ll have paid $200 for that small window in lost heat and another $200 in air conditioning.

A cheap solution is to add curtains, shades, or plastic of some sort. These should not be placed too close to the window, or you won’t have a decent air gap, nor so far that the air will not be static in the gap. For small gaps between the window glass and your plastic or curtain, the heat transfer rate is proportional to the thermal conductivity of air, k, and inversely proportional to the air gap distance, ∂.

Q = ∆T A k /∂.

R  = ∂/k.

The thermal conductivity of air, k, is about .024 BTU/ft. hr°F. We thus confirm that the the R-value for an air gap of 9/16” or 1/20 foot is about 2 in these units. Though the typical air gap between the glass is less, about .3″ there is some more stagnant air outside the glass an that counts towards the 9/16″ of stagnant air. The k value of glass or plastic is much higher than of air, so the layers of glass or plastic add almost nothing to the total heat transfer resistance.

Because the R value of glass and plastic is so low, if you cover your window with a layer of plastic sheet that touches the window, the insulation effect is basically zero. To get insulation value you want to use a gap between about ½” and 1” in thickness. If you already have a 2 paned window of R value 2, you can expect to be able to raise your insulation value to 4 by adding a plastic sheet or single curtain at 9/16” from the glass.

Sorry to say, you can’t raise this insulation value much higher than 4 by use of a single air gap that’s more than 1″ thick. When a single gap exceeds this size, the insulating value drops dramatically as gas circulation in the gap (free convection) drives heat transfer. That’s why wall insulation has fiber-glass fill. For your home, you will want something more attractive than fiberglass between you and the window pane, and typical approaches  include cellular blinds or double layer drapes. These work on the same principle as the single sheet, but have extra layers that stop convection.

My favorite version of the double drapes is the federalist version, where the inner drape is near transparent, shim cloth hangs close to the window, with a heavier drape beyond that. The heavier curtain is closed at night and opened in daytime; where insulation is needed, the lighter cloth hangs day and night. This looks a lot better than a roll-type window shade, or bamboo screen. Besides, with a roll-shade or bamboo, you must put it close to the window where it will interfere with the convection flow, that is cold shedding from the shut window.

Another nice alternative is a “cell shade” These are folded lengths of two or more stiff cloths that are formed into honeycombs ½” to 2” apart. This empty thickness provides the insulating power of the shade. Placed at the right distance from the window, the cell shade will add 3 or more to the overall R value of the window (1/12 ft / .024 BTU/ft. hr°F = 3.5 ft2hr°F/BTU). As with a bamboo screen, all this R value goes away if the shade is set at more than about 1” from the window or an interior shade. At a greater thickness that this, the free convection flow of cold air between the window and the shade dominates, and you get a puddle of cold air on the floor. 

I would suggest a cellular shade that opens from the bottom only and is translucent. This provides light and privacy; a shade that is too dark will be left open. Behind this, my home has double-pane windows (when I was single the window was covered by a layer of plastic too). The see-through shade provides insulation while allowing one to see out the window (or let light in) when the shade is drawn. You want to be able to see out; that’s the reason you had a window in the first place. Very thick, insulating curtains and blinds seem like a waste to me – they are enough thicker to add any significant R-value, they block the light, and if they end up far from the window, the shedding heat loss will more than offset any small advantage from the thick cloth.

One last window insulation option that’s worth mentioning is a reflective coating on the glass (an e-coating). This is not as bad an idea as you might think, even in a cold climate as in Detroit. A surprising amount of heat tends to escape your windows in the form of radiation. That is, the heat leaves by way of invisible (infra –red) light that passes unimpeded through the double pane glass. In hot climates even more heat comes in this way, and a coating is even more useful to preserve air conditioning power. Reflective plastic coats are cheap enough and readily available, though they can be hard to apply, and are not always attractive.

You can expect to reduce the window heat loss by a factor of 3 or more using these treatments, reducing the heat loss through the small window to $1.00 or so per month, far enough that the main heat loss is through the walls. At that point, it may be worth putting your efforts elsewhere. Window treatments can save you money, make a previously uninhabitable room pleasant, and can help preserve this fair planet of ours. Enjoy.

Updated, Feb 9, 2022, REB.