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Re: [nycjunto-discuss] alternative energy

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  • Perry E. Metzger
    ... You do, he does not. Here is Kirchoff s law: At thermal equilibrium, the emissivity of a body (or surface) equals its absorptivity. So lets analyze the
    Message 1 of 19 , Sep 30, 2005
      Bob Armstrong <bob@...> writes:
      >> Imagine a white metal sphere and a black metal sphere out in space at
      >> about our distance from the earth. One half of each sphere is
      >> illuminated by the sun and absorbs energy. One half is facing the rest
      >> of the universe and is radiating it out by black body
      >> radiation. Assume good heat transfer within the sphere. If both
      >> radiate out freely on the same curves, but one absorbs more slowly,
      >> then you end up with the one that absorbs more slowly at a lower
      >> asymptotic temperature. Again, you haven't considered the entire
      >> system.
      >
      > That is exactly the ultimate experiment I propose and will approximate .
      > Kirchoff says they will be the same asymptotic temp and so do I .

      You do, he does not.

      Here is Kirchoff's law:

      At thermal equilibrium, the emissivity of a body (or surface)
      equals its absorptivity.

      So lets analyze the situation, shall we? Since you keep invoking poor
      Mr. Kirchoff, we'll use his law over and over.

      Assume that we have two thin plates in space, one meter square, aimed
      with one side squarely facing the sun, and the other side facing
      space, at the same distance from the sun. For purposes of simplicity,
      assume that space is at 0K instead of 3K -- it makes little difference
      other than making the calculation simpler. Lets also assume the
      distance from the sun is 1AU, so we can just use the earth's solar
      constant.

      We arrange for the front of one plate to be a perfect black body, and
      for the the front of the other to reflects 90% of the incident
      radiation. Let us further assume (for simplicity) that the rear of
      each plate is a perfect black body (i.e. is painted exactly like the
      front of the first plate). This will simplify our calculation.

      The first plate is a perfect black body and therefore absorbs all
      incident radiation. At equilibrium, therefore, it will be absorbing
      1367W of energy. Based on Kirchoff's law, once we achieve thermal
      equilibrium, the object must also be radiating 1367W of energy. The
      surface area of the object is about 2m^2 (consider both the front and
      the back), so it has to be emitting 683.5W/m^2.

      The thermodynamic temperature of a black body can be measured by its
      radiation by the Stefan-Boltzmann law, which states that the radiant
      flux is equal to the Stefan-Boltzmann constant times the fourth power
      of the temperature. Its value is 5.6704E-8.

      So we divide 683.5 by 5.670400E-8 and take the fourth root. I'll save
      you the trip to a calculator and note the value is 331.3K.

      Now, lets consider the plate with one side 90% reflective and one side
      a perfect black body. Total absorption is, by definition, 1367/10 or
      136.7W. Total emission, by Kirchoff's law, must again be 136.7W. We
      have to now emit from both sides -- but the corollary of Kirchoff's
      law, we emit only 1/10th as much through the side that is 90%
      reflective*, so we have the black body side emitting 124.3W/m^2 or
      so. (We can't measure temperature directly from the other side because
      it is not a black body -- it is emitting about 12.4W/m^2 or so).

      We now note that by the Stefan-Boltzmann law, a black body emitting
      124.3W/m^2 is at a temperature of 216.4K.

      Please note that 331K and 216K are NOT the same temperature. (By the
      way, for the Kelvin and Celsius challenged, 331K is about 136F, and
      and 216K is about -71F -- different temperatures indeed.)

      (*that's a slight lie -- no substance reflects in all bands equally,
      so the emission in the much colder temperature band will be different
      (and likely higher, resulting in an even lower temperature), but we'll
      ignore that for this exercise, just as we ignored the fact that there
      are no perfect black bodies.)

      Anyway, if you have better math to present, please tell me about
      it. Feel free to correct anything I did wrong.

      > Otherwise you could transfer energy laterally ( not front to back )
      > between them making a heat engine . That doesn't compute .

      Er, these objects CAN act as a heat engine, if you like. There is
      nothing wrong with that -- it doesn't violate the laws of
      thermodynamics. Heat engines are just fine in an environment where the
      heat is moving from a high heat spot (the sun) to a low heat region
      (the entire universe).

      >>>>First, we need to know the number we're starting with. The solar
      >>>>constant is 1367W/m^2.
      >>>
      >>> Of course , the solar constant is not 4 or even 3 digits constant .
      >>> In fact at least half and estimate range to .8 and .9 and greater
      >>> of total variance in mean earth temperature has been found explained
      >>> by variation in solar radiance . I claim it will eventually be
      >>> understood to be 1.00 .
      >>
      >>
      >> I have my doubts on that. That would assume that there is no
      >> greenhouse effect at all, and there very clearly is. That would also
      >> assume there is no variance in albedo, and there very clearly is.
      >
      > Do a little web surfing or go to my http://cosy.com/views/warm.htm for
      > links . The variability in solar output and it's correlation with
      > earth's temperature is an established fact .

      So what? The question is whether there is a 100% correlation between
      variance in the earth's temperature and solar radiance, not whether
      there is a majority correlation.

      > I used to think quite a bit about albedo because there are such extreme
      > changes in it . How can the earth not get trapped in an ice age when
      > that positively feedsback increases in the reflectivity of the planet
      > so much ?

      Large scale environmental models are good ways of examining that problem.

      > I have come to believe that the energy density at this distance
      > from the sun is the only determinant of mean temperature ,

      We all have our personality quirks.

      If you would like to bet money on this, even after I showed you the
      math, please, by all means, bet me on it. We will need to come up with
      an adequate statement of what we are betting on, of course.

      >>> Of course in the same 24 hours an average nuke will produce 24,000
      >>> MWh versus your 3MW .
      >>
      >> 3MWh is one acre. The average nuclear installation covers many many
      >> acres. If it covers a mere 8 acres (not unreasonable including
      >> security buffers etc) you have the same output.

      I note that you didn't catch my math error here, and it was a huge
      one-- 24GWh vs 24MWh. Still, 8,000 acres isn't really that much
      space. The world is filled with wastelands.

      Consider also that there is plenty of room in orbit, and you get both
      an extra 25% (the amount of sunlight blocked by the atmosphere and
      clouds), and the sun shines 24x7, and you can aim perpendicularly at
      the sun at all times in orbit. Moving the power to the ground via
      microwaves loses maybe 20% but you've more than made up for that here
      already...

      >> Also, we *will* ultimately produce multiple absorber panels, which
      >> will have much higher efficiency still.
      >
      > I see your .3 efficiency being reached in labs . 0.2 seems to be
      > about SoA commercially .

      At the moment, yes, because of cost. However, manufacturing techniques
      only improve with time, and with them, costs fall.

      > But the talk is of doubling the lab efficiencies to .6 or more .

      Not yet really. No one knows how to make a multiple absorber -- yet.

      > That would change the area required to equal a moderate nuke from
      > 19 to 6 square miles .

      My number is 12.5 square miles for a "worst case" 24GWh/day setup --
      the latitude of NY in wintertime. Nearer the equator is much
      better. If we do get to .6 (multiple absorbers), we'd need about 6.25
      square miles at that "worst case" latitude.

      >> Consider also that
      >>
      >> a) This is just a way of collecting energy. We can store it and
      >> transport it in a variety of ways.
      >> b) The price will only go down with time, and the efficiency will only
      >> go up.
      >
      > Pumping water is hard to beat .

      At the moment, yes. There are better things on the horizon. We can
      also move power around very efficiently over extreme distances if
      we're willing to use superconducting cables.

      --
      Perry E. Metzger perry@...
    • Bob Armstrong
      There are other things in life than these list jousts , illuminating as they may sometimes be ; some take precedence . In your example , which I applaud as an
      Message 2 of 19 , Oct 5, 2005
        There are other things in life than these list jousts , illuminating
        as they may sometimes be ; some take precedence .

        In your example , which I applaud as an excellent exercise , you
        analyze a very different situation than that which we were
        discussing : balls with uniform surface properties . Equivalently ,
        rotating spheres like the earth . Your example does point out one effect
        that should affect earth temperature : change in albedo over the
        poles as opposed to equatorial regions and would be , in fact , a
        negative feedback on the expansion of ice caps . That's my unanalyzed
        intuition in any case . Confirm or disconfirm if you like .

        Perry E. Metzger wrote:
        > Bob Armstrong <bob@...> writes:
        >>> Imagine a white metal sphere and a black metal sphere out in space at
        >>> about our distance from the earth. One half of each sphere is
        >>> illuminated by the sun and absorbs energy. One half is facing the rest
        >>> of the universe and is radiating it out by black body
        >>> radiation. Assume good heat transfer within the sphere. If both
        >>> radiate out freely on the same curves, but one absorbs more slowly,
        >>> then you end up with the one that absorbs more slowly at a lower
        >>> asymptotic temperature. Again, you haven't considered the entire
        >>> system.
        >> That is exactly the ultimate experiment I propose and will approximate .
        >> Kirchoff says they will be the same asymptotic temp and so do I .
        >
        > You do, he does not.
        >
        > Here is Kirchoff's law:
        >
        > At thermal equilibrium, the emissivity of a body (or surface)
        > equals its absorptivity.
        >
        > So lets analyze the situation, shall we? Since you keep invoking poor
        > Mr. Kirchoff, we'll use his law over and over.
        >
        > Assume that we have two thin plates in space, one meter square, aimed
        > with one side squarely facing the sun, and the other side facing
        > space, at the same distance from the sun. For purposes of simplicity,
        > assume that space is at 0K instead of 3K -- it makes little difference
        > other than making the calculation simpler. Lets also assume the
        > distance from the sun is 1AU, so we can just use the earth's solar
        > constant.
        >
        > We arrange for the front of one plate to be a perfect black body, and
        > for the the front of the other to reflects 90% of the incident
        > radiation. Let us further assume (for simplicity) that the rear of
        > each plate is a perfect black body (i.e. is painted exactly like the
        > front of the first plate). This will simplify our calculation.
        >
        > The first plate is a perfect black body and therefore absorbs all
        > incident radiation. At equilibrium, therefore, it will be absorbing
        > 1367W of energy. Based on Kirchoff's law, once we achieve thermal
        > equilibrium, the object must also be radiating 1367W of energy. The
        > surface area of the object is about 2m^2 (consider both the front and
        > the back), so it has to be emitting 683.5W/m^2.
        >
        > The thermodynamic temperature of a black body can be measured by its
        > radiation by the Stefan-Boltzmann law, which states that the radiant
        > flux is equal to the Stefan-Boltzmann constant times the fourth power
        > of the temperature. Its value is 5.6704E-8.
        >
        > So we divide 683.5 by 5.670400E-8 and take the fourth root. I'll save
        > you the trip to a calculator and note the value is 331.3K.
        >
        > Now, lets consider the plate with one side 90% reflective and one side
        > a perfect black body. Total absorption is, by definition, 1367/10 or
        > 136.7W. Total emission, by Kirchoff's law, must again be 136.7W. We
        > have to now emit from both sides -- but the corollary of Kirchoff's
        > law, we emit only 1/10th as much through the side that is 90%
        > reflective*, so we have the black body side emitting 124.3W/m^2 or
        > so. (We can't measure temperature directly from the other side because
        > it is not a black body -- it is emitting about 12.4W/m^2 or so).
        >
        > We now note that by the Stefan-Boltzmann law, a black body emitting
        > 124.3W/m^2 is at a temperature of 216.4K.
        >
        > Please note that 331K and 216K are NOT the same temperature. (By the
        > way, for the Kelvin and Celsius challenged, 331K is about 136F, and
        > and 216K is about -71F -- different temperatures indeed.)
        >
        > (*that's a slight lie -- no substance reflects in all bands equally,
        > so the emission in the much colder temperature band will be different
        > (and likely higher, resulting in an even lower temperature), but we'll
        > ignore that for this exercise, just as we ignored the fact that there
        > are no perfect black bodies.)
        >
        > Anyway, if you have better math to present, please tell me about
        > it. Feel free to correct anything I did wrong.
        >
        >> Otherwise you could transfer energy laterally ( not front to back )
        >> between them making a heat engine . That doesn't compute .
        >
        > Er, these objects CAN act as a heat engine, if you like. There is
        > nothing wrong with that -- it doesn't violate the laws of
        > thermodynamics. Heat engines are just fine in an environment where the
        > heat is moving from a high heat spot (the sun) to a low heat region
        > (the entire universe).
        >
        >>>>> First, we need to know the number we're starting with. The solar
        >>>>> constant is 1367W/m^2.
        >>>> Of course , the solar constant is not 4 or even 3 digits constant .
        >>>> In fact at least half and estimate range to .8 and .9 and greater
        >>>> of total variance in mean earth temperature has been found explained
        >>>> by variation in solar radiance . I claim it will eventually be
        >>>> understood to be 1.00 .
        >>>
        >>> I have my doubts on that. That would assume that there is no
        >>> greenhouse effect at all, and there very clearly is. That would also
        >>> assume there is no variance in albedo, and there very clearly is.
        >> Do a little web surfing or go to my http://cosy.com/views/warm.htm for
        >> links . The variability in solar output and it's correlation with
        >> earth's temperature is an established fact .
        >
        > So what? The question is whether there is a 100% correlation between
        > variance in the earth's temperature and solar radiance, not whether
        > there is a majority correlation.
        >
        >> I used to think quite a bit about albedo because there are such extreme
        >> changes in it . How can the earth not get trapped in an ice age when
        >> that positively feedsback increases in the reflectivity of the planet
        >> so much ?
        >
        > Large scale environmental models are good ways of examining that problem.
        >
        >> I have come to believe that the energy density at this distance
        >> from the sun is the only determinant of mean temperature ,
        >
        > We all have our personality quirks.
        >
        > If you would like to bet money on this, even after I showed you the
        > math, please, by all means, bet me on it. We will need to come up with
        > an adequate statement of what we are betting on, of course.
        >
        >>>> Of course in the same 24 hours an average nuke will produce 24,000
        >>>> MWh versus your 3MW .
        >>> 3MWh is one acre. The average nuclear installation covers many many
        >>> acres. If it covers a mere 8 acres (not unreasonable including
        >>> security buffers etc) you have the same output.
        >
        > I note that you didn't catch my math error here, and it was a huge
        > one-- 24GWh vs 24MWh. Still, 8,000 acres isn't really that much
        > space. The world is filled with wastelands.
        >
        > Consider also that there is plenty of room in orbit, and you get both
        > an extra 25% (the amount of sunlight blocked by the atmosphere and
        > clouds), and the sun shines 24x7, and you can aim perpendicularly at
        > the sun at all times in orbit. Moving the power to the ground via
        > microwaves loses maybe 20% but you've more than made up for that here
        > already...
        >
        >>> Also, we *will* ultimately produce multiple absorber panels, which
        >>> will have much higher efficiency still.
        >> I see your .3 efficiency being reached in labs . 0.2 seems to be
        >> about SoA commercially .
        >
        > At the moment, yes, because of cost. However, manufacturing techniques
        > only improve with time, and with them, costs fall.
        >
        >> But the talk is of doubling the lab efficiencies to .6 or more .
        >
        > Not yet really. No one knows how to make a multiple absorber -- yet.
        >
        >> That would change the area required to equal a moderate nuke from
        >> 19 to 6 square miles .
        >
        > My number is 12.5 square miles for a "worst case" 24GWh/day setup --
        > the latitude of NY in wintertime. Nearer the equator is much
        > better. If we do get to .6 (multiple absorbers), we'd need about 6.25
        > square miles at that "worst case" latitude.
        >
        >>> Consider also that
        >>>
        >>> a) This is just a way of collecting energy. We can store it and
        >>> transport it in a variety of ways.
        >>> b) The price will only go down with time, and the efficiency will only
        >>> go up.
        >> Pumping water is hard to beat .
        >
        > At the moment, yes. There are better things on the horizon. We can
        > also move power around very efficiently over extreme distances if
        > we're willing to use superconducting cables.
        >

        --
        Bob Armstrong -- http://CoSy.com -- 719-337-2733
        NoteComputing Environment : http://CoSy.com/CoSy/
        WTC vision : http://CoSy.com/CoSy/ConicAllConnect/
        Liberty : http://cosy.com/Liberty/
      • Perry E. Metzger
        ... The earth doesn t have a simple uniform reflectivity in the sense that the atmosphere is not equally transparent to all wavelengths, and energy from the
        Message 3 of 19 , Oct 5, 2005
          Bob Armstrong <bob@...> writes:
          > In your example , which I applaud as an excellent exercise , you
          > analyze a very different situation than that which we were
          > discussing : balls with uniform surface properties . Equivalently ,
          > rotating spheres like the earth .

          The earth doesn't have a simple uniform reflectivity in the sense that
          the atmosphere is not equally transparent to all wavelengths, and
          energy from the sun arrives at a different wavelength than the earth's
          black body radiation re-radiates it. A number of gases have the
          property of being fairly transparent to visible light and relatively
          less transparent to IR radiation. That's why greenhouse gases are
          important.

          As for spheres of uniform color vs. plates, in my example, I chose to
          analyze plates one side of which was a perfect black body because
          doing the original problem (spheres of uniform white and uniform
          black) is much harder to analyze. I invite you to try the more
          complicated problem, however, if you insist. I'm not going to bother
          -- my point has been made.

          By the way, I think you might have the grace to explicitly admit that
          your claims about the laws of thermodynamics are wrong. You very
          clearly claimed that Kirchoff's laws imply the equilibrium temperature
          of a body at a given distance from the sun does not depend on its
          reflectivity, and I very clearly demonstrated that you were wrong.


          Perry
        • Bob Armstrong
          ... The crux of your demonstration is the non-uniformity of reflectivity toward and away from the source of heat . As I said , I was amiss to not be clear that
          Message 4 of 19 , Oct 6, 2005
            Perry E. Metzger wrote:
            > Bob Armstrong <bob@...> writes:
            >> In your example , which I applaud as an excellent exercise , you
            >> analyze a very different situation than that which we were
            >> discussing : balls with uniform surface properties . Equivalently ,
            >> rotating spheres like the earth .
            >
            > By the way, I think you might have the grace to explicitly admit that
            > your claims about the laws of thermodynamics are wrong. You very
            > clearly claimed that Kirchoff's laws imply the equilibrium temperature
            > of a body at a given distance from the sun does not depend on its
            > reflectivity, and I very clearly demonstrated that you were wrong.

            The crux of your demonstration is the non-uniformity of reflectivity
            toward and away from the source of heat . As I said , I was amiss to
            not be clear that I was only thinking about objects with uniform
            reflectance or alternatively ones averaged by rotation .

            --
            Bob Armstrong -- http://CoSy.com -- 719-337-2733
            NoteComputing Environment : http://CoSy.com/CoSy/
            WTC vision : http://CoSy.com/CoSy/ConicAllConnect/
            Liberty : http://cosy.com/Liberty/
          • Bonnie
            I kept meaning to get back to this thread, and am finally doing it because I just saw a relevant blog posting to pass along. See:
            Message 5 of 19 , Jan 15, 2006
              I kept meaning to get back to this thread, and am finally
              doing it because I just saw a relevant blog posting to pass
              along.

              See:
              http://geddesblog.blogspot.com/2006/01/baloghblog-storing-summer-solar-heat.html

              That comment of Bob's kept resonating in my head as the days
              got shorter and therefore, colder: That the sun doesn't even
              give us enough energy to keep the ground from freezing for
              the winter. How true! (Despite the temporary January thaw we
              just got here.)

              Anyway, that blog posting mentions:

              1. Capturing the winter cold, in the form of pykrete, to cool
              buildings in the summer
              2. Borehole thermal energy storage (BTES), an in-ground heat
              sink for seasonal energy storage -- and a link to a Canadian
              community setting this up to heat a 52-house subdivision in
              winter

              As for my yurt:
              1) My solar panels have arrived, I've got my batteries,
              but haven't done anything with them yet. But I'm not
              trying to heat the place with them, just provide a little
              light and music now and then!
              2) For heating the yurt, I'm going to take advantage of
              some natural solar energy storage...in the form of downed
              trees on the property I'll be burning!

              Bonnie
            • Bob Armstrong
              ... Bore hole heat sinks are becoming a common feature in new construction in NYC . You will find the caps of the 1600 bore holes in front of all the Sciame
              Message 6 of 19 , Jan 16, 2006
                Bonnie wrote:
                > I kept meaning to get back to this thread, and am finally
                > doing it because I just saw a relevant blog posting to pass
                > along.
                >
                > See:
                > http://geddesblog.blogspot.com/2006/01/baloghblog-storing-summer-solar-heat.html
                >
                > That comment of Bob's kept resonating in my head as the days
                > got shorter and therefore, colder: That the sun doesn't even
                > give us enough energy to keep the ground from freezing for
                > the winter. How true! (Despite the temporary January thaw we
                > just got here.)
                >
                > Anyway, that blog posting mentions:
                >
                > 1. Capturing the winter cold, in the form of pykrete, to cool
                > buildings in the summer
                > 2. Borehole thermal energy storage (BTES), an in-ground heat
                > sink for seasonal energy storage -- and a link to a Canadian
                > community setting this up to heat a 52-house subdivision in
                > winter

                Bore hole heat sinks are becoming a common feature in new
                construction in NYC . You will find the caps of the 1600' bore
                holes in front of all the Sciame reconstructed buildings along
                Front St and Peck Slip by the Seaport .

                > As for my yurt:
                > 1) My solar panels have arrived, I've got my batteries,
                > but haven't done anything with them yet. But I'm not
                > trying to heat the place with them, just provide a little
                > light and music now and then!
                > 2) For heating the yurt, I'm going to take advantage of
                > some natural solar energy storage...in the form of downed
                > trees on the property I'll be burning!

                Likewise here in CO . We've got years of wood to burn just
                to clean up the property , especially because of beetle killed
                trees . Our main need to enhance our solar efficiency is thermal
                curtains to cover at night the picture windows which collect a
                lot of sun in the afternoon .

                We have some solar powered lights ( LED ) out on the front
                gate , and would love to get a solar powered remote gate
                opener . I got an ad for a solar powered freezer which seems
                like a neat use because you get the power just when you need
                it - winter takes care of itself . On some of the most remote
                and severe curves between here and Denver , there are solar
                powered warning flashers . These are the really practical uses
                of a few square feet of solar panels ( a few watts ) and will
                clearly dominate its use in temperate climates for a long time .

                --
                Bob Armstrong -- http://CoSy.com -- 719-337-2733
                CoSy 2005 NewsLetter & MidWinter Party[ 19 ]
                : http://CoSy.com/home.htm#20051231
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