Quote (Thor123422 @ 8 Feb 2021 21:53)
According to google 1 kg of uranium makes 24 million kWh of power at a nuclear plant.
Converted to joules that's 8.64 x 10^13 J, or 8.64 x10^7 MJ
Sending something to space takes 109 MJ/KG, so we want 700KG so 76300 MJ. So we've got a wasted energy content of 0.08% of the total energy.
Even if my numbers are off, I've got a factor of over 1000 to make up for it.
It's not really the energy content that's the problem, as much as it's just the practical aspects of sending things to space.
According to wikipedia (
https://en.wikipedia.org/wiki/Nuclear_power_in_the_United_States ), the electricity produced by nuclear energy in the United States per year is around 809 terrawatt-hours, or around 2.912 x 10^ 12 MJ.
According to the DoE (
https://www.energy.gov/ne/articles/5-fast-facts-about-spent-nuclear-fuel ), the U.S. produce around 2000 metric tons of used fuel per year.
2 x 10^6 kg * 700 * 109 MJ/kg = 1.526 x 10^ 11 MJ required in raw energy to send the waste itself to space, which corresponds to 5.24% of the produced energy just for the payload-part of the shoot-to-the-moon operation. Hence, there is not enough buffer to disregard the other factors in the equation.
Specifically, keep in mind that the launch vehicle itself also has to be brought into orbit, meaning that the self-weight to payload ratio must not exceed 19. Now, to look at one example, Saturn V (
https://en.wikipedia.org/wiki/Saturn_V ), the record holder among heavy lift launch vehicles, has a weight of around 2900 tons and can carry a payload of 140 tons, for a self-weight to payload ratio of 20.7.
In other words: using this method, sending the nuclear waste to space with the conventional measures would require roughly as much energy as it produced. And that's not even factoring in the additional conversion loss during the production of the rocket fuel.
Let me know if you find an error in my math.
This post was edited by Black XistenZ on Feb 8 2021 05:17pm