[ExI] Space Based Solar Power vs. Nuclear Fission

[hkhenson]

At 10:44 PM 5/17/2008, Lee wrote:
>Keith wrote in (English-speaking Google News....)
>
> > If the coupling between energy and food is as modeled then the ride
> > down the back side of peak oil/energy is going to reduce the world's
> > human population to a small fraction of the current level...
> > ...
> > The rational thing would be to develop replacement energy, and space
> > based solar power seems to be the best bet.
>
>Perhaps you've addressed this before, but I missed it.
>
>Having described the high initial investment costs regarding
>space based solar power, why exactly do you believe this
>to be a more feasible solution than further development of
>conventional nuclear power?

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.

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