The problem is available on this link on page 91: http://books.google.ca/books?i
ID: 2092081 • Letter: T
Question
The problem is available on this link on page 91: http://books.google.ca/books?id=zLV-G4-fgHAC&pg;=PA91&lpg;=PA91&dq;="estimate+the+mean+emissivity"&source;=bl&ots;=aJ9T35gXRh&sig;=bHNdiuWw_0Xo0N13dxo69DRteSA&hl;=fr&sa;=X&ei;=V0BHUdfvKqvA4AOz9YCgCQ&ved;=0CDIQ6AEwAQ#v=onepage&q;="estimate the mean emissivity"&f;=false (Sorry for long link). The question demands us to estimate the mean emissivity from a figure of brightness temperature fluctuation over a space and time plan in the atmosphere. I've been searching everywhere on the Internet for help for this problem but still cannot see how to estimate the atmosphere mean emissivity from this figure. Anyone can help? Thanks a lot. (Solution for this problem isn't available in the book, I own it and I know it's not in it)Explanation / Answer
THIS WILL BE HELPFUL FOR YOU Okay. Let's jump right in. The flux density (energy throughput per unit area) going into the Earth's climate system is: F = (S / 4) (1 - A) [1] where S is the Solar constant and A the albedo. The Solar constant is the amount of sunlight coming through a unit area at Earth's mean distance from the Sun. According to recent satellite estimates, this averaged 1,366 watts per square meter over the period 1951-2000. The albedo is the "bolometric Bond albedo"--the fraction of energy falling on a planet, averaged over all wavelengths, that gets reflected back out to space. A for Earth is about 0.306 according to NASA. Given that equation and those figures, F = 237 watts per square meter for the Earth. Where did the figure of 1/4 come from? Well, Earth receives sunlight on its two-dimensional cross-sectional area (?R2), but its surface area (4?R2) is three-dimensional. The 1-A comes about because the fraction of incoming sunlight not reflected is the fraction absorbed, and it's the amount absorbed that goes into the climate system. Conservation of energy says the Earth must radiate out as much energy as it receives, at least in the long run. Inverting the Stefan-Boltzmann law, we can use our figure for F to calculate the Earth's "radiative equilibrium temperature" (also called effective temperature or emission temperature): Te = (F / (? ?))0.25 [2] Here, ? is the Earth's emissivity, its radiation efficiency compared to that of a perfect radiator (or "black body"). ? is assumed to be 1 for a planetary atmosphere radiating into space. ? is the Stefan-Boltzmann constant. This constant has the value 5.6704 x 10-8 W m-2 K-4 in the SI. F = 237, ? = 1 and ? as above gives Earth an equilibrium temperature of Te = 254 K. Now, since water freezes at 273 K, we can see that if sunlight were the whole story, the Earth would be frozen over. The surface temperature is actually about 288 K, and the extra 34 K comes from the greenhouse effect, created largely by water vapor and carbon dioxide in the Earth's atmosphere. How to calculate the size of the greenhouse effect on a planet? It's difficult. There are many, many factors that affect a planet's surface temperature. Sunlight and greenhouse gases are the big ones, but one must also account for clouds, convection, evaporation of seawater, absorption of sunlight in the atmosphere, etc., etc., etc. What you really should do is write a "radiative-convective model" of the Earth's atmosphere, taking into account as many of these processes as possible. But that's a tedious and complicated undertaking. We can get an approximate answer with a "gray approximation" for the greenhouse gases and "parameterizations" for the other effects. Part II--Greenhouse Gases A body is considered "gray" if it reacts the same way to light at all wavelengths. A "blackbody," as mentioned above, is one that absorbs and emits light perfectly; it's a special case of the gray approximation. We will actually use a "semigray" approximation--we will assume the atmosphere and the gases in it react the same way to light in the infrared spectral range, which is the range where radiation from the Earth and the atmosphere is important. This is usually delimited as being from about 4 to 125 microns wavelength. If you know the "gray infrared optical thickness" for an atmosphere, you can calculate a raw "greenhouse temperature" for the planet's surface: T0 = Te (1 + 0.75 ?)0.25 [3] What is optical thickness? It measures the extent to which a beam of light is diminished going through a material body. A high optical thickness means very little light gets through (it is scattered or absorbed); a low optical thickness means a lot gets through. Optical thickness is dimensionless--no units--and it is defined so that a body with an optical thickness of ? = 1 diminishes the intensity of a beam of light by a factor of e (2.718...). For a more detailed explanation of optical thickness, try the Optical Thickness Page. For infrared light, scattering can usually be neglected (this is not quite true for Venus). We will assume all extinction is due to absorption. Curve-fitting the results of radiative-convective models indicates that the optical thickness contributed by a greenhouse gas tends to follow the square root of the gas's partial pressure. Partial pressure is defined as: Pi = fi P [4] where Pi is the partial pressure of gas i, fi its volume fraction, and P the overall atmospheric pressure. This is not the pressure you would find from gravitational mechanics if the gas existed in isolation. For more on that, see the Partial Pressure Page. The greenhouse effect on Earth is caused mainly by water vapor and carbon dioxide. The same gases are important in the greenhouse effect on Venus and Mars, with carbon dioxide taking the lion's share. We can check our gray greenhouse model by seeing if it gets the right results for Venus and Mars as well as the Earth. Here are some characteristics of the three planets: Parameter Venus Earth Mars Solar constant S (W m-2): 2611.0 1366.1 589.1 Bond albedo A (no units): 0.750 0.306 0.250 Climate flux F (W m-2): 163 237 110 Equilibrium temp. Te (K): 232 254 210 Surface temp. Ts (K): 735 288 214 Surface pressure P (Pa): 9,210,000 101,325 636 CO2 fraction (no units): 0.965 0.000332 0.9532 H2O fraction (no units): 0.000030 0.00387 0.000210 For theoretical reasons which will be explored later, we know that the Earth's raw greenhouse temperature would be about 317 K. Back-calculating through equation [3], we can predict that Earth's infrared optical thickness should be about ? = 1.90. The total optical thickness of the atmosphere is just the sum of the partial optical thicknesses contributed by each greenhouse gas. From equation [4], the partial pressures of CO2 and H2O in Earth's atmosphere should be about 33.6 and 392 pascals, respectively. Partial optical thickness should be proportionate to the square root of partial pressure, and those figures would be 5.80 and 19.8, respectively. One more datum is needed--how strong a greenhouse gas is carbon dioxide compared to water vapor? The tables of Houghton (2002) imply that water vapor is about three times as strong, on average. We can therefore fill in the proportionality constants in these equations for partial optical thickness as follows: ?CO2 = kCO2 PCO20.5 [5] ?H2O = kH2O PH2O0.5 [6] With what we know so far, kCO2 = 0.029 and kH2O = 0.087. Let's follow through with the gas calculations for all three planets, and then use them to find the raw greenhouse temperature: Parameter Venus Earth Mars PCO2 (Pa): 8,890,000 33.6 606 PH2O (Pa): 280 392 0.134 ?CO2 (no units): 86.5 0.168 0.714 ?H2O (no units): 1.51 1.72 0.031 ? (no units): 87.9 1.89 0.745 T0 (K): 664 317 235 Part III--Mechanisms which Cool the Planetary Surface Our raw greenhouse temperatures from the table above are too low for Venus and too high for Earth and Mars. Our figure for Venus, although in the right ballpark, is not going to improve. The sulfuric acid clouds have a huge effect on Venus's surface temperature, and we're not accounting for them. But we can improve the figures for Earth and Mars by taking into account cooling mechanisms. Aside from its own thermal radiation, three major mechanisms cool Earth's surface at the expense of the atmosphere. Taking the values from the Earth radiation budget of Trenberth et al. (2009), these mechanisms are: absorption of sunlight by the atmosphere: 78.2 W m-2 sensible heat (conduction and convection): 17 W m-2 latent heat (evaporation of seawater): 80 W m-2 The previous version of this page listed Earth's direct "window" radiation loss of 40 W m-2 as a cooling mechanism for Earth's surface. It is, but it shouldn't count separately from Earth's radiation in general. The major mechanism by which Earth's surface cools itself is by radiating about 371 watts per square meter, of which about 331 goes into the atmosphere and 40 to space. But that radiation loss will directly relate to Earth's temperature through the Stefan-Boltzmann radiation law. Radiation to space should never have counted as a separate cooling mechanism. The table above should list only the cooling mechanisms that work apart from Earth's own radiation. But at any rate, we can't just put the same figures down for Venus and Mars, because conditions are different there. We have to come up with plausible "parameterizations"--simple functions which give us believable figures. Absorption of sunlight. By definition, this goes up with optical thickness. From disparate sources in the literature, the solar radiation which actually makes it to the planetary surface averages 16.8 watts per square meter for Venus, 161.4 W m-2 for Earth and 102 W m-2 for Mars. A judicious bit of curve-fitting allows us to relate the resulting visual optical thicknesses ?VIS to the infrared ? values found above: ?VIS = 0.36 (? - 0.723)0.411 [7] Of course this means that for ? < 0.723, we have to simply assign ?VIS = 0 to avoid imaginary numbers, but no planet in our sample has an atmospheric optical thickness this low. Our algorithm for atmospheric cooling loss is then, simply, Beer's Law: Labs = F - F exp(-?VIS) [8] This next section introduces a lot of new terms, so bear with me. We now consider sensible heat and latent heat. These can be lumped together as the surface "convective flux" Fc. Lorenz and McKay (2003) suggest that this can be parameterized with an equation of the form: Fc = Fsi ? / (C + D ?) [9] where Fsi is the surface solar illumination and C and D are constants. Fsi is defined as: Fsi = F - Labs [10] I think there is a conceptual problem here; there is no reason the heat input for convection should be confined to solar heating. I would replace Fsi above with the total radiative flux absorbed by the surface, which we might call Fabs: Fabs = (1 - As) Fsi + ? (F0 - F) [11] Here, As is the surface albedo--0.108, 0.15, and 0.25 for Venus, Earth and Mars, respectively. ? is the infrared emissivity of the surface, usually taken to be about 0.95 on average by most global climate models. By Kirchhoff's Law, absorptivity = emissivity for an object in local thermodynamic equilibrium. F0 is the raw greenhouse climate flux, found from the raw greenhouse temperature T0 (see above) by using the Stefan-Boltzmann radiation law: F0 = ? ? T04 [12] We will also need a proportionality constant in our Fc equation, to account for the facts that we are using Fabs and not Fsi, and that we want to match Earth's observed convective flux (sensible flux of 17 W m-2 + latent flux of 80 W m-2 = 97 W m-2). Our equation for the convective flux therefore becomes, finally, Fc = 0.369 Fabs ? / (-0.6 + 2 ?) [13] Part IV--Applying the Model Now we have all we need for a complete model. From the flux from the raw greenhouse temperature, we merely subtract the losses to find the flux from the surface: Fs = F0 - Labs - Fc [14] And that gives us our final surface temperature estimate, using, for the third time in this analysis, the Stefan-Boltzmann law: Ts = (Fs / (? ?))0.25 [15] Here's what we get for each planet: Parameter Venus Earth Mars Labs (W m-2): 146 75.6 7.9 Fc (W m-2): 1821 94 39.5 F0 (W m-2): 10,500 544 164 Fs (W m-2): 8533 374.4 116.6 T (K): 631 289 216 Comparing these model-predicted temperatures to the actual figures, we have a fairly decent match: Parameter Venus Earth Mars T (K): 631 289 216 Ts (K): 735 288 214 To take the Pearson product-moment correlation coefficient (r) between the predicted and actual values would give misleading results--there are only three data points and the model is incredibly tweaked. But just for the heck of it, r = 0.9994. Ta da!Related Questions
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