[ExI] Space Based Solar Power vs. Nuclear Fission

[Stefano Vaj]

On Sun, May 18, 2008 at 10:53 PM, hkhenson <hkhenson at rogers.com> wrote:
> It may not be.  If we could build a space elevator, the energy
> payback time is incredibly short--at least for the energy needed to
> lift a power sat to GEO.

Speaken of unproven, or rather non-existing, technologies... :-)

Having said that, needless to say I am absolutely in favour of
investigating and implementing both satellite- or moon-based solar
energy and researching the fundamentals that might lead to a space
elevator.

Speaking of lowest hanging fruits, I do not see such things as very
rapid-deployment solutions, however. Please note however that I am
much better informed of the state-of-the art of nuclear fusion.

Stefano Vaj

> Specific orbital energy is u/2r, (398,600/42,000)/2 or -4.75Mj/kg
>
> Potential is -9.5Mk/kg and kinetic is 4.75Mj/kg
>
> Potential at the earth's surface is -62.6 MJ/kg; the difference is 53.1Mj/kg.
>
> Using a space elevator, the rotation of the earth provides the
> kinetic energy.  Since a joule is a watt-second; 53,100
> kW-s/kg/3600kW-s/kWh is 14.75 kWh/kg
>
> A kW/kg power sat repays its lift energy 14 hour and 45 minutes after
> being turned on.  A 2kg/kW power sat would take 29.5 hours.
>
> There is a heck of a lot of sunlight out there, and you don't need
> much structure to capture it in zero g.  Whatever rate you get for
> paying off solar cells on the ground, they will repay it at least
> three times faster in orbit.
>
> With rockets, it's still not too bad.
>
> http://www.ilr.tu-berlin.de/koelle/Neptun/NEP2015.pdf
>
> Neptune is about 3 times the capacity of a Saturn 5, and this design
> was done by some of the same people so it's solid engineering.
>
> This vehicle delivers 350 mt to LEO, and 100 mt to lunar orbit or
> GEO. To lift 100 mt to GEO Neptune uses 3762-mt of propellant for the
> first stage, 1072 mt second stage and 249 mt for the third totaling 5077 mt.
>
> SSME O2 to H2 ratio is 6 to
> 1.
> http://www.pw.utc.com/vgn-ext-templating/v/index.jsp?vgnextrefresh=1&vgnextoid=75a0184c712de010VgnVCM100000c45a529fRCRD
>
> I.e., 1 part in 7 of the propellant is LH. or about 725 mt of
> LH.  The launch site would make electrolytic hydrogen out of water
> (the only long term source). That costs about 50 kWh/kg plus another
> 15 kWh to liquefy the H2.  Add 5 kWh for liquefying oxygen at 6 x .8 kWh/kg.
>
> That would be 70 MWh per mt, or 70 GW hours for 1000 tons, or 50.8
> GWh for 725 mt. At a kg/kW, 100 tons of satellite produces 100,000
> kW, or 0.1 GW.  Thus it would take 508 hours to pay back the lift
> energy or 21.2 days, 42.4 days for 2kg/kW.
>
> Rocket efficiency here would be 14.75/508 or 2.9%.
>
> A nuclear tug stage shuttling between LEO and GEO might double this
> efficiency raising the payload from 100 mt to 200 mt.  That is the
> consequences of a 15 km/sec exhaust velocity.
>
> It's the high cost of rocket hardware, not lift energy payback, that
> makes power sats expensive.  If we could get that down . . .
>
> As for conventional nuclear power, I can't see it being built fast
> enough.  The need for replacement energy as oil fades out of the
> picture is around a GW/day.  That's in addition to the problem of
> neutrons being diverted to make plutonium 239 out of DU.
>
> Keith
>
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