11 ms·

> 14 milligram HB11 can produce 300 kWh energy That's an enormous amount of energy from a minuscule amount of mass. Using their predictions, I could run my ho

> 14 milligram HB11 can produce 300 kWh energy That's an enormous amount of energy from a minuscule amount of mass. Using their predictions, I could run my house for a year on less than 1 gram of HB11. How does this compare with the energy output of current-day fission reactors?

sathackr 9y ago
To answer my own question, using numbers from [1]: U235 = 79,390,000 MJ/kg 1 kWh = 3.6MJ 1kg u235 = ~22,000,000kWh 1g u235 = ~22,000 kWh of energy For the HB11 reaction: 1000 / .014 = ~21,000 kWh/g So about the same as current fission. Except without the nasty byproducts. [1] https://www.quora.com/How-many-milligrams-of-uranium-is-needed-to-produce-one-googol-kilowatt-hours-of-electricity https://www.quora.com/How-many-milligrams-of-uranium-is-need... edit: complete failure of math.
philipkglass 9y ago
With mainstream present-day fission technology, it takes about 200 tonnes of natural mined uranium to fuel a 1000 MWe reactor for a year: http://www.world-nuclear.org/information-library/nuclear-fuel-cycle/introduction/nuclear-fuel-cycle-overview.aspx http://www.world-nuclear.org/information-library/nuclear-fue... The same 300 kWh would then require about (300 / (1000 * 24 * 365 * 1000)) * (200000000) = 6.85 grams of natural uranium. Note that this is well below the theoretical energy density available from the fission of uranium. The main loss is because the naturally fissile isotope of uranium, U-235, is less than 1% of naturally occurring uranium. The much more common U-238 can be used completely in a breeder reactor, but there are currently only two operating breeder reactors in the world that generate electricity: https://en.wikipedia.org/wiki/BN-600_reactor https://en.wikipedia.org/wiki/BN-600_reactor https://en.wikipedia.org/wiki/BN-800_reactor https://en.wikipedia.org/wiki/BN-800_reactor

sathackr 9y ago

To answer my own question, using numbers from [1]: U235 = 79,390,000 MJ/kg 1 kWh = 3.6MJ 1kg u235 = ~22,000,000kWh 1g u235 = ~22,000 kWh of energy For the HB11 reaction: 1000 / .014 = ~21,000 kWh/g So about the same as current fission. Except without the nasty byproducts. [1] https://www.quora.com/How-many-milligrams-of-uranium-is-needed-to-produce-one-googol-kilowatt-hours-of-electricity https://www.quora.com/How-many-milligrams-of-uranium-is-need... edit: complete failure of math.

philipkglass 9y ago

With mainstream present-day fission technology, it takes about 200 tonnes of natural mined uranium to fuel a 1000 MWe reactor for a year: http://www.world-nuclear.org/information-library/nuclear-fuel-cycle/introduction/nuclear-fuel-cycle-overview.aspx http://www.world-nuclear.org/information-library/nuclear-fue... The same 300 kWh would then require about (300 / (1000 * 24 * 365 * 1000)) * (200000000) = 6.85 grams of natural uranium. Note that this is well below the theoretical energy density available from the fission of uranium. The main loss is because the naturally fissile isotope of uranium, U-235, is less than 1% of naturally occurring uranium. The much more common U-238 can be used completely in a breeder reactor, but there are currently only two operating breeder reactors in the world that generate electricity: https://en.wikipedia.org/wiki/BN-600_reactor https://en.wikipedia.org/wiki/BN-600_reactor https://en.wikipedia.org/wiki/BN-800_reactor https://en.wikipedia.org/wiki/BN-800_reactor