Average Lighting and Temperature
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Rama
physics Average Lighting and Temperature |
In this section we want to know if Rama can provide a similar temperature and lighting as on Earth, at least in theory. More precisely we are interested in the average temperature and in the average luminance, which measures the amount of visible light. These two quantities are not independent because the visible light is partially absorbed by the ground, which gets warmer because of this. Thus, the temperature induced by the lighting alone could be larger than the desired temperature, which would be a serious obstacle to the feasibility of Rama (the opposite case is not a problem, because we can easily add heat sources to increase the temperature without increasing the amount of visible light). The main goal of this section is to know whether this is the case or not.
Although Rama is much smaller than the Earth, it is still a very complex system with an atmosphere, an "ocean", several linear light sources, etc. Computing the "exact" temperature in Rama would thus require a very complex model, similar in complexity to the wheather or climate models for the Earth. However, the average temperature of the Earth can be estimated with a very simple analytic model using many simplifying hypotheses. Since Rama is supposed to provide the same lighting and temperature as on Earth, we assume here that these hypotheses also hold for Rama. We then use them to check whether or not Rama can provide a similar temperature and lighting as on Earth.
The next section presents the simplifying hypotheses that can be made to estimate the average temperature of the Earth, and how they are used to compute it. We then apply them to the case of Rama in the last section.
The temperature of the Earth depends mainly on two factors: how much lighting power it receives from the Sun, and how this light interacts with the Earth. Basically, light can interact with matter in 3 ways:
In general all 3 cases happen at the same time, i.e. some part is reflected, some part is absorbed, and the remaining part is transmitted. The ratio of light which is reflected is the albedo, noted here $\alpha$. The ratio of light which is absorbed is the absorptivity. A perfect mirror has albedo 1 and absorptivity 0. A black body has albedo 0 and absorptivity 1, i.e. it absorbs everything, hence its name.
The amount of light which is emitted by thermal radiation is maximal for a black body, and is equal to $\sigma T^4\,W.m^{-2}$, where $\sigma=5.67\,10^{-8}\,W.m^{-2}.K^{-4}$ is the Stefan-Boltzmann constant, and $T$ the temperature in Kelvins. For any other material thermal radiation is smaller than this. The ratio between the actual thermal radiation and the black body radiation is the emissivity, noted here $\epsilon$. The Kirchhoff's law of thermal radiation states that emissivity and absorptivity are equal.
Finally, it should be noted that albedo, absorptivity and emissivity all depend on wavelength. For instance a green object appears as such because its albedo in the green wavelengths is larger than in the other wavelengths.
The first simplifying hypothesis is that we can reduce the whole range of wavelengths to only two distinct ranges, and replace wavelength dependent quantities by their average in these two ranges. These two ranges, that we call short wave and long wave, are those corresponding to the power spectrums of black bodies at the Sun and Earth temperatures, $5800\,K$ and $288\,K$ respectively. They are indeed well separated, at about $4\mu m$ (see Fig. 1). Note that the short wave range contains the visible and the near-infrared ranges.
The second simplifying hypothesis is that the following properties hold:
The third simplifying hypothesis is that all the physical quantities, which in reality vary in space and time, can be replaced with their space and time average. This includes:
The fourth and final simplifying hypothesis is that the thermal radiation of the Earth surface can be approximated with the one of a black body, i.e. that it is equal to $\sigma T_s^4$. Note that the surface is obviously not a true black body (otherwise its albedo would be 0), but this approximation is still quite accurate. Note also that, as a consequence of this hypothesis (following from Kirchhoff's law), we can assume that the surface absorbs all the long wave radiation it receives.
From the above hypotheses, we can immediately compute the average surface temperature if there was no atmosphere. Indeed, in this case, the surface receives $S_0/4$ watts per square meter, reflects $\alpha_e S_0/4$, emits $\sigma T_s^4$, and that's all (see Fig. 2). At termal equilibrium the net flux must be 0, which gives $S_0/4-\alpha_e S_0/4-\sigma T_s^4=0$, from which we get $T_s=[(1-\alpha_e)S_0/4\sigma]^{1/4}=255\,K=-18°C$.
Taking the atmosphere into account gives two equations for the two unknowns $T_s$ and $T_a$ and we get in this case (see Fig. 3 and The idealized greenhouse model): $T_s=[(1-\alpha_e)S_0/(4\sigma-2\epsilon)]^{1/4}$.
For $\epsilon=0.78$ we get $T_s=288\,K=15°C$ (and $T_a=242\,K=-31°C$), which is the measured average surface temperature of the Earth. We can see from this that the partial absorption and re-emission of the infrared radiations by the atmosphere, known as the greenhouse effect, plays a critical role for the surface temperature. Without it this average surface temperature would be $-18°C$ instead of $15°C$.
We can also see that if the greenhouse effect was at its maximum, i.e. if the atmosphere was absorbing and re-emitting all the infrared emitted by the surface ($\epsilon=1$), the average surface temperature would be $30°C$, which is way too hot. Yet, this is precisely what happens in Rama: no visible or infrared radiation inside Rama can directly escape into space. All radiations must eventually be absorbed by the shell, where they are re-emitted as infrared, half into space, half into Rama. This is why it is not immediately obvious whether or not Rama can provide similar lighting and temperature conditions as on Earth.
We would like to get "similar" temperature and lighting in Rama as on Earth. For the temperature, this means $T_s=288\,K=15°C$. For the lighting, more precisely for the illuminance, things are a bit more complex. On Earth, the illuminance when the Sun is at the zenith, with a clear sky, is about $100.000\,lux$. But this is really bright, and we can assume that $70.000\,lux$ is sufficient for Rama (this corresponds to the Sun at $45°$ from the zenith). We would also like a day and night cycle in Rama, such that the average illuminance is half its maximum value, i.e. $35.000\,lux$.
On Earth, at the ground level, $1000\,W.m^{-2}$ of incoming solar radiation gives an illuminance of about $100.000,lux$. But this illuminance comes from the visible light alone, which represents only about 45% percent of the incoming energy [1]. For Rama, we can study two extreme scenarios:
As explained in the introduction, we assume here that the simplifying hypotheses that can be used to compute the average Earth surface temperature can also be used for Rama. Thus we assume that we can consider the short and long wave ranges separately, that all quantities can be averaged, that the Rama surface can be appoximated with a black body, etc. To simplify things even further, we will also suppose that Rama is an infinitely long and infinitely thin cylinder.
Now, let's assume that the linear light sources emit $S$ watts per meter along the axis, in the short wave range. Whatever the values of the atmosphere absorptivity / emissivity $\epsilon$ and the surface albedo $\alpha$, no radiation can escape directly into space, and so all of it must eventually be absorbed and re-emitted by the shell. Thus, at thermal equilibrium the infrared absorbed and re-emitted by the shell must be $S$ watts per meter. This power is $2\pi r_s\sigma T_s^4$, where $r_s$ is the radius of Rama's surface. We thus get directly $T_s=[S/2\pi r_s\sigma]^{1/4}$, which is independent of the atmosphere and ground properties.
We can now consider the lighting inside Rama. In the short wave range, the surface eventually absorbs all the power emitted by the sources (since it can't escape), and nothing more. We can thus ignore the atmosphere and conclude directly that the surface absorbs $I_0=S/2\pi r_s$ watts per meter in average. Since it has albedo $\alpha$ and no transmittance, its absorption ratio is $1-\alpha$. This means that the ground receives $I=I_0/(1-\alpha)$ in average, and reflects $\alpha I=\alpha I_0/(1-\alpha)$ right away.
In the long wave range, the shell emits $\sigma T_s^4$ towards the inside. Like in the short wave range, the surface must eventually absorb all this power, which can't escape directly into space, and nothing more. We can therefore ignore the atmosphere again, and conclude directly that the shell absorbs $\sigma T_s^4$ in average. Finally, the shell also emits $\sigma T_s^4$ towards space, which corresponds to $I_0$, the net flux of short wave radiation (and, since the outer surface is larger than the inner surface, its temperature is lower). All these fluxes are summarized in Fig. 4.
The results are that
We can now apply the above model to our two above scenarios. In the "incandescent" scenario $I=350\,W.m^{-2}$, which gives $T_s=269\,K=-4°C$. In the "energy-saving" scenario $I=157\,W.m^{-2}$, which gives $T_s=220\,K=-52°C$. In both case the average temperature is too low, but this can easily be fixed by adding heat sources. The conclusion is that Rama can indeed provide a similar temperature and lighting as on Earth.
In fact we don't need to add heat sources, i.e. to produce more energy, to increase the temperature. Instead, we can partially reflect the infrared emitted into space, which is wasted, back towards the shell. One way to do this is to add a second shell around the first one, close to it but with an empty, insulating space between the two (see Fig. 5).
In this case we easily get $T_s=[2I_0/\sigma]^{1/4}=[2(1-\alpha)I/\sigma]^{1/4}$ and $T_s=320\,K=47°C$ in the "incandescent" scenario, and $T_s=262\,K=-11°C$ in the "energy-saving" scenario. The temperature increase is too large for the "incandescent" scenario, and not large enough for the "energy-saving" scenario. But this can be fixed by making holes in the outer shell to reduce its green house effect or, conversely, by adding a third shell around the first two. This shows that we can build a composite shell emitting $\Delta$ times more radiation towards the inside than towards the outside, with $\Delta$ adjustable. The ground temperature is then $T_s=[\Delta I_0/\sigma]^{1/4}$. From this we find that, in order to get $T_s=288\,K=15°C$, it suffice to use have $\Delta=1.3$ in the "incandescent" scenario, or $\Delta=2.9$ in the "energy-saving" scenario.
4 Conclusion
Applied to Rama, the most basic simplified model used to computed the average temperature of the Earth shows that the temperature deriving from Rama's light sources is less than the desired temperature. This is true whatever the light sources, provided their power is such that it provides the same illuminance as on Earth on Rama's surface. But this model also shows that, by using composite shells, one can increase the temperature to the desired level without requiring more power and without increasing the illuminance. It is thus possible to provide the same average lighting and temperature in Rama as on Earth.
In the following we will use the two scenarios below, which both have an albedo $\alpha=0.15$, an average illuminance of $35.000\,lux$ and an average temperature of $288\,K=15°C$:
As a concluding remark, note that in the most energy efficient scenario the light source power at the surface is $2I_0=2*133\,W.m^{-2}$ at its maximum value during the day and night cycle. Multiplying this by the inner surface of Rama $A_s\approx 2\pi r_sl_s+2\pi r_s^2=2412\,km^2$, we get the total power which is needed in this case to heat and illuminate Rama: $2A_sI_0=620\,GW$, i.e. about 170% of the total power of all the nuclear reactors on Earth in 2011 (440 reactors for $369\,GW$ [2])!