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January 27, 2010

What Would it Take for a Yottawatt Civilization

There is a new website yottawatts from Thorium, which prompted this post to answer how in general to get yottawatts from fission, fusion or solar

Here are the units prefix definitions from wikipedia.

10^3 W kW kilowatt 
10^6 W MW megawatt 

10^9 W GW gigawatt  (current large hydro have 1 to 20 gigwatts. Nuclear and coal plants can have 1-2 gigawatts )

10^12 W TW terawatt (The total power used by humans worldwide (about 16 TW in 2006) is commonly measured in this unit.)

10^15 W PW petawatt (the total energy flow of sunlight striking Earth's atmosphere is estimated at 174 PW)

10^18 W EW exawatt 

10^21 W ZW zettawatt 

10^24 W YW yottawatt (the Sun outputs approximately 386 Yottawatts)

A yottawatt civilization would be roughly equivalent to a Kardashev 1.75 civilization. Kardashev level One is able to access and use all of the energy of the sun that strikes the earth. Kardashev Level Two is able to use all of the energy of the sun.



Power Sources Super Solar, Fusion and Deep Burn Fission

Previously this site has discussed deep burn nuclear fission and using Uranium from the Ocean. Deep burn would mean using nuclear fuel about 100 times more efficiently than we do now. Burning all of the Uranium, Plutonium or thorium completely. There would only be waste material that has half life of less than 30 years left.

With deep burn and maximum mining of uranium and thorium we could get 1 million times more nuclear fission power than we do now. 100 times more efficiency and 10,000 times more material. 600 million tons per year.

So scaling up from about 400 gigawatts up to 400 petawatts. This rate of usage would go through the uranium in the ocean in about 6 years and would be going through the uranium and thorium in the crust at very good clip.

There is an estimated 40 trillion tons of Uranium and 120 trillion tons of thorium in the Earth's crust.

So it would take about 267,000 years to go through all of the uranium and thorium at the 400 petawatt rate.

There is an estimated of 600 trillion tons or 12 times the amount in the earth's crust for Uranium not in the Sun in the solar system.

So going up to 1 zettawatt would use up the fissionable nuclear material in about 1000 years. Going up to 1 yottawatt would use it up in one year.
If the ratio of thorium to uranium held in the solar system as it does for the crust then there would be about 4 years of yottawatt power using all of the thorium and uranium that is not in the sun.

Nuclear Fusion

Nuclear fusion appears to be on the verge of a breakthrough

About 1 in 6500 hydrogen atoms in seawater is deuterium. Deuterium abundance on Jupiter is about 2.25×10^−5 (roughly 22 atoms in a million, or 15% of the terrestrial deuterium-to-hydrogen ratio.

There is enough water in the ocean to provide energy for 3 X 10^11 years at the current rate of energy consumption. The availability of Lithium on land is sufficient for at least 1000 if not 30000 years, and the cost per kWh would be even smaller than that of Deuterium. If the oceans is included it is estimated that there is enough fuel for 3 X 10^7 years.

16 Terawatts X 3 X 10^11 years OR
16 Petawatts X 3 X 10^8 years OR
16 Exawatts X 3 X 10^5 years OR
16 Zettawatts X 300 years OR 1 Zettawatt for 4800 years OR
1 Yottawatt for 4.8 years

This is only using the ocean. Using Jupiter for nuclear fusion fuel would enable billions of years of yottawatt energy.

UPDATE -
A reader made the detailed calculations while I had made some assumptions in a rough calculation which were not correct.

Jupiter's upper atmosphere is composed of about 88–92% hydrogen. The interior contains denser materials such that the distribution is roughly 71% hydrogen. Mass 1.8986×10^27 kg, 317.8 Earths or 1/1047 of the Sun.

1.9 * 10^27 * 22* 10^-6 = 4.18*10^22 kg = 4.18 X 10^19 tons X.9 = 3.6 X10^19

Deuterium ratio of the solar system. Higher ratio of deuterium in Uranus and Neptune than Jupiter and Saturn

Gaseous Planets (Deuterium parts per million)
21 ± 8 H2 Jupiter (spectroscopic) =1
26 ± 7 H2 Jupiter (MS in situ) =1
15–35 H2 Saturn (spectroscopic) =1
65 (+2.5/–1.5) H2 Neptune (spectroscopic) 2.6
55 (+35/–15) H2 Uranus (spectroscopic) 2.2



hydrosphere of earth 1.5×10^18 short tons
hydrogen 1.67X10^17
deuterium 2.5X10^13 tons

Here is a posting on another site with an attempt to calculate sustainability of Kardashev civilizations

Europa's oceans of 3 × 10^18 m3, slightly more than two times the volume of Earth's oceans.

Uranus (14.5 X 2.2 earth masses), Neptune 17 X (2.6 higher deuterium ratio) earth masses,
Saturn 95 earth masses,
171 earth mass equivalents with Jupiter Deuterium ratio
So maybe 10 million years of Deuterium fusion

The advanced civilization would have to figure out how to make proton-proton fusion (regular hydrogen) work if they wanted to make non-solar fusion last. Proton-proton fusion does occur in the Sun.

Proton-proton fusion is not one of the reactions that is considered a candidate for planetary based nuclear fusion However, if you are not planetary based and have enough technology then you probably can find a way.

Super solar Energy

As noted by the definitions, capturing and using 0.3% of the energy from the sun enables a yottawatt civilization.

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