Solar-pumped laser power transmission, a way to dramatically decrease launch costs?
in [Physics]
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From: William Mook on 16 Feb 2010 14:55 On Feb 15, 9:48 pm, OM <o...(a)sci.space.history> wrote: > On Mon, 15 Feb 2010 16:52:55 -0800 (PST), "Scott M. Kozel" > > <koze...(a)comcast.net> wrote: > >I can see that you have put a lot of thought into this > > ...Most of it the result of years of substance abuse, especially > sniffing paint fumes from a paper bag while high on really poorly made > acid. > > OM > > -- > > ]=====================================[ > ] OMBlog -http://www.io.com/~o_m/omworld [ > ] Let's face it: Sometimes you *need* [ > ] an obnoxious opinion in your day! [ > ]=====================================[ That's libel OM. Naughty naughty.
From: Alain Fournier on 16 Feb 2010 21:01 Pat Flannery wrote: > Temperature at 4.1 million miles is 2,160F according to the above > article, so assuming halving the distance increases the temperature four > times over (I think that's how it works, though the large solar radius > may screw this up; it's going to at least double) that makes the temp it > could be facing over 8,000 F. Yes the large solar radius does screw this up. The correct way to compute this is to measure the solid angle that the sun covers at a given distance. For something far away, when you double the distance the solid angle is divided by four. But when you are close to the sun the the squaring law no longer applies. Very close to the surface of the sun, the temperature will basically be constant, because whether you are 100 km from the surface or 200 km makes little difference, the sun basically fills half the sky in both cases. Alain Fournier
From: William Mook on 17 Feb 2010 10:01 On Feb 16, 2:49 pm, David Spain <nos...(a)127.0.0.1> wrote: > William Mook <mokmedi...(a)gmail.com> writes: > > I believe my experience exceeds yours at this point. I understand > > that you are clueless. That does not mean I am. > > You seem to understand little, I await your SPS on orbit. > > Dave Alright.
From: William Mook on 17 Feb 2010 10:39 On Feb 16, 9:01 pm, Alain Fournier <alain...(a)sympatico.ca> wrote: > Pat Flannery wrote: > > Temperature at 4.1 million miles is 2,160F according to the above > > article, so assuming halving the distance increases the temperature four > > times over (I think that's how it works, though the large solar radius > > may screw this up; it's going to at least double) that makes the temp it > > could be facing over 8,000 F. > > Yes the large solar radius does screw this up. The correct way to compute > this is to measure the solid angle that the sun covers at a given distance. > For something far away, when you double the distance the solid angle > is divided by four. But when you are close to the sun the the squaring > law no longer applies. Very close to the surface of the sun, the temperature > will basically be constant, because whether you are 100 km from the surface > or 200 km makes little difference, the sun basically fills half the sky > in both cases. > > Alain Fournier This is correct when adjusting for energy. Inverse law doesn't strictly apply because the sun subtends some area. Another way of saying this is that the inverse square curve doesn't go to infinity at zero radius. It goes to the temperature of the solar surface at the solar surface. This makes the energy rise less than the inverse square would suggest - and the correction SUBTRACTS from that curve. (infinity is bigger than any large finite number) At 3.5 million km its something like 27 degrees of the sky. The entire sky covers 4 pi steradians. Think about a ball that has a unit radius. It has an area of 4 pi square units. Now a circle with a radius of 13.5 degrees on the surface of this sphere 13.5 / 57.3 = 0.24 units radius - having an area of 0.18 square units (pi*0.24^2) which is 1.43% of the entire sky. So, there is a slight adjustment at 3.5 million km from the sun, but not much - due to the fact that the light is coming from a sensible surface not a point at zero radius. This diffuses the light. The big issue is though, that even at 3.5 million km the inverse square law, which over-estimates energy levels, provides only a slight over-estimate. In the end, this nuance should not be confused with the fact that in a vacuum energy is exchanged by radiant transfer. So, radiant energy in, versus radiant energy out. Imagine a gray shell of titantium 1,128.4 mm in diameter that absorbs perfectly 50% of energy incident on it, and reflects away 50% of the energy incident on it sitting at a point that has an energy flux of 2 million watts per square meter. This ball has a projected area of 1 square meter and a total surface area of 4 square meters. It reflects away 1 million watts of energy and absorbs 1 million watts of energy which raises its temperature to the point where it glows at an intensity where that 1 million that gets absorbed is radiated away as well. Since the entire surface is glowing, and 1 million watts is coming in then 1 million watts must be going out to balance it. Since the entire area is 4 square meters, the flux per unit area is 250,000 watts/m2 What must this temperature be? Well, Stephan Boltzmann tells us j* = 5.67e-8 * T^4 Re-arranging T = (j*/5.67e-8)^(0.25) = 1,449.1 Kelvin Which is well below the melting point of Titanium. If the Titanium shell is 10 mm thick, we can see that the shell has a volume of 40 liters and a mass of 181.6 kg. And with a heat capacity of 597.5 J/kg-K we can see that it will take no less than 2.4 minutes to reach that temperature at this distance. (imagining that it is removed from behind a highly reflective screen) depending on rate of rotation. This is how its done. You need to know the rate of energy coming into each unit of projected area. You need to know the reflectivity of those surfaces. You then have to calculate the temperature using Stephan Boltzman to see what the temperature must be to balance the heat load. You then look at the temperature, the materials, and their physical properties to get an idea of what will happen in that location. Numerical modelling is best, especially with ray tracing to get accurate heat flows. For example, consider a panel with hairlike fibers on the back side, that are fairly efficient heat pipes. It turns out that with fairly efficient geometries, you can have pretty cool systems even with high heat flows. Add in the ability to reflect wavelengths and to operate refrigeration cycles to radiant emitters (with hair) and you can see that challenging as this engineering is, its perfectly amenable to careful analysis and creative thought. To achieve remarkable results. A started this conversation detailing how two 10 ton satellites, one in GEO and one 3.5 million km from the sun, can displace ALL the nuclear power plants in the USA for less than the cost of TWO nuclear power plants. Once the system is up and running, and proven, 20 satellite pairs can displace ALL power plants in the USA for the cost of four more nuclear power plants. 140 satellite pairs can displace ALL PRIMARY ENERGY USE IN THE WORLD TODAY! for the cost of four more power plants (the power of mass production) So, for the same cost as 10 new nuclear plants, the USA using its existing space faring infrastructure could supply 100% of the world's energy from space, and continue to support agressive industrial expansion thereafter - earning $2 trillion per year - and growing in 30 years to $30 trillion while the world economy grows from $70 trillion today to $1 quadrillion. Of course, if 20% of the revenue goes into expanding our space faring infrastructure, this would be $400 bililion per year - we could do other amazing things - if we combine the skill and creativity to think of them.
From: William Mook on 17 Feb 2010 11:15
On Feb 17, 1:28 am, Pat Flannery <flan...(a)daktel.com> wrote: > Alain Fournier wrote: > > Yes the large solar radius does screw this up. The correct way to compute > > this is to measure the solid angle that the sun covers at a given distance. > > For something far away, when you double the distance the solid angle > > is divided by four. But when you are close to the sun the the squaring > > law no longer applies. Very close to the surface of the sun, the > > temperature > > will basically be constant, because whether you are 100 km from the surface > > or 200 km makes little difference, the sun basically fills half the sky > > in both cases. > > What makes it tricky to figure out is that this isn't just a case of the > object in orbit being bombarded by the sunlight and heating up that way, > but it actually being in the tenuous outer atmosphere of the Sun. > At its hottest points the corona can get up to between 8 and 20 million > degrees Kelvin. > > Pat Well, this isn't tricky at all. What you are saying is that you are NOT in a perfect vacuum. You have some sort of conduction going on from a hot body to a cooler one. Its still the same heat balance equation. So, you add to the radiant transfer the conductive transfer. Now, you're still in pretty much a vacuum. That's important to remember. You're not in a water bath at 8 million degrees. You're in a vacuum where the particles that strike your vehicle are at 8 million degrees. How many? That's the key question. Because we'll assume for arguments sake that anything at 8 million degrees will pretty much give up all its energy to whatever it hits. This isn't precisely accurate, but accurate enough for engineering work. Alright, so, how mch? The parts of the Sun above the photosphere are referred to collectively as the solar atmosphere. They can be viewed with telescopes. There are five principal zones: 1) the temperature minimum, 2) the chromosphere, 3) the transition region, 4) the corona, and 5) the heliosphere. The heliosphere, which may be considered the tenuous outer atmosphere of the Sun, extends outward past the orbit of Pluto to the heliopause, where it forms a sharp shock front boundary with the interstellar medium. The chromosphere, transition region, and corona are much hotter than the surface of the Sun. The reason has not been conclusively proven; evidence suggests that Alfvén waves may have enough energy to heat the corona. The coolest layer of the Sun is a temperature minimum region about 500 km above the photosphere, with a temperature of 4,100 K. This part of the Sun is cool enough to support simple molecules such as carbon monoxide and water, which can be detected by their absorption spectra. Above the temperature minimum layer is a layer about 2,000 km thick, is the chromosphere. The temperature in the chromosphere increases gradually with altitude, ranging up to around 20,000 K near the top. Above the chromosphere there is a thin (about 200 km) transition region in which the temperature rises rapidly from around 20,000 K in the upper chromosphere to coronal temperatures closer to one million kelvins. The temperature increase is facilitated by the full ionization of helium in the transition region, which significantly reduces radiative cooling of the plasma. So, you can see how important radiative transport of energy is. The transition region does not occur at a well-defined altitude. Rather, it forms a kind of cloud layer around chromospheric features such as spicules and filaments, and is in constant, chaotic motion - like weather on Earth - only a lot hotter! The transition region is not easily visible from Earth's surface, but is readily observable from space by instruments sensitive to the extreme ultraviolet portion of the spectrum. The corona is the extended outer atmosphere of the Sun, which is much larger in volume than the Sun itself. The corona continuously expands into the space forming the solar wind, which fills all the Solar System. So, saying that the satellite is in the corona at 3.5 million km doesn't really bother me - and it shouldn't bother you. All you must do is figure out what the heat balance is to get an accurate read on temperature. The low corona, which is very near the surface of the Sun, has a particle density around 1e+11 particles per cubic meter at 700,000 km from the center - the Earth's atmosphere has 2e+25 by comparison. Density falls off as the inverse square of distance once the solar wind is established, but that doesn't occur until we reach the base of the heliosphere at 15 million km.. So, at 3.5 million km the effect is dominated by an exponential function called density lapse rate. At the high end density would be around 4e+9 particles per cubic meter. At the low end, density would be about 1e+7 particles per cubic meter. The average temperature of the corona and solar wind is about 1 million2 million kelvins, however, in the hottest regions it is 8 million20 million kelvins as stated. We are dealing with the average. The heliosphere, which is the cavity around the Sun filled with the solar wind plasma, extends from approximately 20 solar radii (0.1 AU) to the outer fringes of the Solar System. Its inner boundary is defined as the layer in which the flow of the solar wind flows faster than the speed of Alfvén waves. Turbulence and dynamic forces outside this boundary cannot affect the shape of the solar corona within, because the information can only travel at the speed of Alfvén waves. The solar wind travels outward continuously through the heliosphere, forming the solar magnetic field into a spiral shape, until it impacts the heliopause more than 50 AU from the Sun. In December 2004, the Voyager 1 probe passed through a shock front that is thought to be part of the heliopause. Both of the Voyager probes have recorded higher levels of energetic particles as they approach the boundary. So that's the information. Now, temperature is a measure of particle energy. The AVERAGE is 1 milion to 2 million K. The particles are largely protons. This makes calculating their average energy easy. A proton at 7.24e+22 Kelvins contains 1 Joule. Its a linear relationship since temperature is a measure of energy. http://physics.nist.gov/cgi-bin/cuu/Value?jk 2 million K = 2e+6 K So, each particle has 2e+6/7.24e+22 = 2.76e-17 J per poton. Even if we were flying at the base of the corona, where density is highest (1e+11) we'd have contained in each cubic meter of space 1e+11 * 2.76e-17 J = 2.76 micro-joules And its likely to be 1/100th this level at the altitude we're flying. To get a power rating we have to plug in the speed of things. Like a care driving down the highway thorugh a snowstorm, the speed of the car dominates all the other speeds - so, our spacecraft flying at 3.5 million km radius is moving at 120,000 m/sec. So, we're getting energy dumped in for each square meter of projected area (along the line of motion) of; 2.,76e-6 J/m3 * Area (1 for our sphere) * 120,000 m/sec = 0.54 watts per square meter. Compare this to the 1 million watts per square meter coming from sunlight. Increasing the temperature to 20 million K increases the rate to 5.4 watts per square meter. Which has very little impact on the temperature - it is a correction factor in a system dominated by radiant energy. |