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Nukes in space

  1. Jun 5, 2009 #1
    So lets say we have an offworld mining colony somewhere in the solar system and we want to build an orbital refinary and fabrication plant to process the ore, for such an application which choice for power would be more practical, solar thermal or nuclear? How large would they need to be?
     
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  3. Jun 5, 2009 #2

    Astronuc

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    Solar thermal. No need for shielding. The solar flux is on the order of 2 kW/m2 in earth's orbit, and one could do a combination of PV and concentrated solar power. Out by Mars, the flux would drop by a factor of ~4.
     
  4. Jun 5, 2009 #3
    So how large would the solar thermal stuff be?
     
  5. Jun 6, 2009 #4

    Astronuc

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    Refer to this - http://en.wikipedia.org/wiki/Solar_constant#Sunlight_intensity_in_the_Solar_System

    Let's pick 1 kW/m2 as the usable amount of sunlight. Then if one wished to have a 1 MW plant, then one needs an area 1000 kW/(1 kW/m2) or 1000 m2 or 31.6 m x 31.6 m or a disk of 35.7 m dia.

    A 100 MW plant would increase those dimensions by a factor of 10.

    This assumes full use of the energy supplied. When considering thermodynamic cycles, one must look at the difference between energy in and energy out, i.e., Tin and Tout. In space, there is no medium for conduction or convection, so any unused thermal energy must be radiated to space.
     
  6. Jun 8, 2009 #5
    I see, so if we decided to use a nuke reactor in space, how large would the radiators need to be in order to cool it properly?
     
  7. Jun 8, 2009 #6
    Collection of insolation in a thermal solar plant is half the equation. The other half has to radiate into space somewhere (given no atmosphere for conduction). But what if there's a surface in the way. Back radiation reduces efficiency, complicating the problem.

    There are other factors. Is the plateform spinning? Is it feasably to power-up and power-down at the rotation frequency?
     
  8. Jun 8, 2009 #7

    Astronuc

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    That would be a function of temperature where the heat flux is determined by the Stefan-Boltzmann equation.

    http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html

    One may assume the background temperature to be ~4K, if one is sufficiently distant from a planet or sun. The radiators plane should be perpendicular to the axis to the peak thermal source, e.g., sun. For a large solar collector, the radiator would be on the backside of the collector.

    One has to do a trade off between thermal efficiency of the thermodynamic side and the radiator temperature, Trad. The Carnot effiency would be proportional to Thot - Trad, and the radiated power would be proportional to Trad4 - 196 K4.
     
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