This morning’s author is a chemical engineer with nearly 50 years of experience. His article is a more technical paper than usual for CommUnity of Minds, but the topic addressed is an important one for humanity’s future.

His conclusion after a careful analysis is that Nuclear Energy obtained by fission is significantly limited. In the best case scenario, it could provide only 28% of power we need for our current energy demand. It could supply this power for a maximum of 1,724 years.

This is a significant energy resource, but it does not solve our energy crisis; and it brings with it, many environmental and security problems. The details follow:

**Donald B. Halcom**

Gertrude Stein once said, “A rose, is a rose, is a rose”. Paraphrasing the old girl, “A Neutron, is a Neutron, is a Neutron” and “A Joule, is a Joule, is a Joule”.

On this planet there is only one naturally occurring fissile element and that is Uranium 235 (U235). Mankind has learned how to make two new fissile elements and these are Uranium 233 (U233) and Plutonium 239 (Pu239). Both of these manmade elements are made by “neutron capture” from the fission of a fissile element (e.g. U235) followed by the spontaneous neutron beta minus decay to form new fissile elements. U238 is converted to Pu239 and Thorium 232 (Th232) is converted to Uranium 233 (U233). Observe that one must sacrifice one fissionable material to make another fissionable material. There is no free lunch. U238 and Th232 are called “fertile” elements. Notice the “odd” and “even” relationships.

The starting amount of U235 initially present and the amounts of U233 and Pu239 that can be formed are of primary interest to mankind. Our future, in part, depends upon them. Electric power generation will be highly influenced by the finite amounts of these fissile materials. Mankind keeps demanding ever increasing amounts of electricity. The purpose of this analysis is to estimate how long “nuclear” materials can supply such power.

**Inventory Time**

**Figure One** shows the availability of Uranium on the planet (this Uranium contains 0.72% U235 and the rest is U238).

As of 2011 there is about 7.3E6 metric tons of recoverable Uranium on the planet. Notice how the price of Uranium recovery is “sky rocketing”. The harder it is to find the greater the expense to harvest.

With 0.72% U235 in the 7.3E6 metric tons of Uranium for 2011, we get:

U235 5.26 E4 metric tons (fissile) = 52,600 metric tons

U238 7.25 E6 metric tons (fertile)

Th232 2.81 E6 metric tons (another fertile element)

Th232/U235 = 53.4 There is much more thorium than U235.

U238/U235 = 143.7 There is even more U238 than Th232.

As you can see the element that is the most crucial is the least available. Everything depends on this element.

**Fuel Recovery**

The spent nuclear fuel as of 2009 is:

Spent fuel 2.4 E5 metric tons with 0.8% U235 and 1.2% Pu239 (240,000 metric tons)

U235 (0.8%) 1,920 metric tons

Pu239 (1.2%) 2,880 metric tons

Here is the most crucial numbers I found on the internet:

Current world spent fuel rate = 10,500 metric tons/year

World number of Nuclear Reactors = 437 (these are mainly slow neutron light water reactors)

World total electric power from all reactors = 372,210 megawatts

As of now there are:

52,600 metric tons of virgin U235

1,920 metric tons of U235 in spent fuel

2,880 metric tons of Pu239 in spent fuel

500 metric tons of weapons grad Pu239

There is one quick way to check the above data for accuracy. This method involves “back calculation” from the megawatts of electricity (372,210 megawatts = 0.372 terawatts) generated by all of the light water nuclear plants on the planet. First, electrical generation from the heat of a nuclear reactor is limited by the entropy of the process. In the real world, one cannot take “random” heat and produce “ordered” electricity without paying a price. The efficiency of such a process is limited to about 34% based upon entropy. Simply stated, it takes 100 joules of thermal heat to produce 34 joules of electrical energy. Sorry folks but that is the way nature works. Now how much thermal energy do we have to spend the get 0.372 terawatts of yearly electrical energy? The answer is 0.372/0.34 = 1.094 terawatts.

Now 1 watt = 1 joule per second and there are 3.15572E7 seconds in a year. This means that we require the following amount of thermal energy in a year from nuclear fission to produce the required electricity.

1.094E12 watts X 3.15572E7 seconds = 3.453E19 joules

The people who know about nuclear fission have measured that for U235, the heat released is 7.28E13 joules per kilogram of reacted U235. How many kilograms of U235 is this? It is 3.453E19/7.28E13 = 474,313 kilograms or 474.3 metric tons of U235 that must disappear per year. Now the next time that someone asks you how much U235 is consumed to run a nuclear electric plant of some fixed capacity, you can tell them. It ain’t that complicated.

Now compare the number presented above with one that I will derive using 10,500 metric tons of spent fuel being generated per year. I did not make this number up, it was published in 2009. I computed the new numbers using more complicated calculations than I used to compute the above 474.3 metric tons per year. The agreement will be spectacular.

**The Future**

Assume:

The **current** spent fuel rate is equal to the **current** yearly feed rate of fresh fuel = 10,500 metric ton/yr.

Fresh fuel has a typical value of 4.5% U235.

Then:

Fresh Fuel In = (0.045X10,500) = 472.5 metric tons of U235/yr into the world’s **current** nuclear reactors.

Spent Fuel Out = (0.008X10,500) = 84.0 metric tons of U235/yr out of the world’s **current** reactors.

(0.012X10,500) = 126.0 metric tons of Pu239/yr out of the world’s **current** reactors

The future requires more than 437 plants, so assume a 5% compounded growth rate of similar plants.

Here is where the presentation gets more technical. I have developed an equation (that can be evaluated on a spreadsheet) to calculate how long the world can run its total number of nuclear plant before the world supply of U235 is exhausted.

I will present it as easily as possible:

Let: U235Zero = the starting **VIRGIN** reservoir of U235 in the world = 52,600 metric tons

U235Time = the amount in the reservoir at any time in metric tons

G = the compound growth rate as a fraction=0 (No growth) or 0.05 for 5% growth (or other rates).

DT = the nominal time scale = 1 year

ConsU235 = the **yearly** consumption rate of **VIRGIN** U235 in metric tons per year

^ = Exponent

Then:

U235Time = U235Zero – ConsU235 *DT* [1 + (1 + G) + (1 + G)^2 + (1 + G)^3 + (1 + G)^4 + …… ]

To solve this equation set U235Time = 0 (this is when the fuel runs out), then:

U235Zero = ConsU235 *DT* [1 + (1 + G) + (1 + G)^2 + (1 + G)^3 + (1 + G)^4 + …… ]

Set: Coeff = ConsU235 / U235Zero

This means that: 1 = Coeff * DT* [1 + (1 + G) + (1 + G)^2 + (1 + G)^3 + (1 + G)^4 + …… ]

Remember that DT = 1 year

When the right side of the above equation reaches 1.00 then the **VIRGIN** U235 is exhausted.

I have created an Excel program that will solve the above equation in short order.

**Calculation Results**

This table shows the results from calculations for various conditions. Zero growth means no new plants.

Method |
Growth % |
ConsU235 |
U235 Exhaustion Time Years |

No recycle U235 |
0 |
472.5 |
112 |

No recycle U235 |
5 |
472.5 |
39 |

Recycle U235 |
0 |
388.5 |
136 |

Recycle U235 |
5 |
388.5 |
42 |

The total **spent** fuel at the end of the exhaustion of all of the U235 would be about 1,428,000 metric tons yielding about 17,136 metric tons of Pu239. This is significantly less than the 52,600 metric tons of U235 currently in the ground. Current light water slow neutron nuclear plants are not long range solutions to the power requirements of the future. This recovered Pu239 plus weapons grade Pu239 could also be used in modified plants somewhere along the time line to extend the exhaustion time a **little**.

There are currently 437 nuclear plants. If we use 5% growth and recycle U235 then at the point of exhaustion the number of plants equals (1.05)^42 * 437 = 3392 plants of the current equivalent sizes. Sounds reasonable? The amount of power being generated at that point would be (1.05)^42 * 372,210 = 2.9E6 megawatts = 2.9 terawatts.

Remember when we were kids back in the late 40s, there was a Saturday morning children’s radio program called “Let’s Pretend”. Well let us return to that era and play **Let’s Pretend**.

Assume that we could **instantly** build 1,000 new Fast Neutron Breeder Reactor plants incorporating metallic U238 or Th232 (fertile). Assume that we use all of the current U235 and Pu239 (fissile) in the world in these 1,000 plants. Further assume that these plants have liquid sodium as coolant with electric power generation equal in every plant. Now in theory we can calculate the power generation by using the fissile fuel in all of the plants and dividing the result by 1,000 to get the power per plant. This is what we will do. Hey, this is **Let’s Pretend**.

Please examine the following titled “Plutonium Breeding Ratio”. Examine especially the part about “doubling time”. Doubling time is the time required to produce “new” fissile fuel equal to the quantity of consumed fissile fuel originally in the breeder reactor. It says that the target is to obtain this value after about ten years of operation. The word “target” means that it is not actually known whether this can be achieved or not. But hey, this is **Let’s Pretend**.

**Plutonium Breeding Ratio**

In the breeding of plutonium fuel in breeder reactors, an important concept is the breeding ratio, the amount of fissile plutonium-239 produced compared to the amount of fissionable fuel (like U-235) used to produce it. In the liquid-metal, fast-breeder reactor (LMFBR), the target breeding ratio is 1.4 but the results achieved have been about 1.2 . This is based on 2.4 neutrons produced per U-235 fission, with one neutron used to sustain the reaction.

The time required for a breeder reactor to produce enough material to fuel a second reactor is called its doubling time, and present design plans target about ten years as a doubling time. A reactor could use the heat of the reaction to produce energy for 10 years, and at the end of that time have enough fuel to fuel another reactor for 10 years. |

Here are the data that we will need to do some more calculations. First of all, here is the world’s **current** consumption of energy. All of this concerns fast neutron breeder reactors with a replacement ratio equal to 1.0.

World’s Current Consumption of Energy = 16.4 Yearly Average terawatts = 5.18E20 joules/year. Includes nuclear, hydro, fossil fuels, solar, wind etc.

Total current **fertile** nuclear fuel: U238 = 7,250,000 metric tons

Th232 = 2,810,000 metric tons (will make U233)

Total Fertile = 10,000,000 metric tons

Total Current **fissile** nuclear fuel: U235 and all Pu239 including weapons grade Pu239 = 58,000 metric tons

An aside is needed here. From the internet, I obtained the following data:

U235 atomic heat of fission = 202.5 Mev (Million electron volts used in fundamental work)

Pu239 atomic heat of fission = 207.1 Mev

For U235 the heat of fission per kilogram = 7.28E13 joules/Kg

But: Pu239perKg/U235perKg = ((207.1)(235))/((202.5)(239)) = 1.0056 = 1

Assume: Heat of fission per Kg U235 = Heat of fission per Kg Pu239 = Heat of fission per Kg U233 = 7.28E13 joules/Kg

This makes life simpler.

Now here comes our next magic simplification. Our **reserve** of unused fuel becomes the unused fertile material and not the unused fissile material. We now have a reserve of 10,000,000 metric tons and not 58,000 metric tons. This is because we “brew” the new fissile material in the breeder reactor to replace the fissile material we just burned. Sounds like magic and in some sense it is. We will repeatedly consume 58,000 metric tons of fissile material until the fertile material supply runs dry.

Now here is where the “doubling time” comes in. We will assume this time to be 10 years as mentioned above. This means that our yearly fuel consumption rate is 5,800 metric tons per year.

With the afore mentioned heat of fission: 5,800,000 Kg X 7.28E13 = 4.22E20 joules/yr

4.22E20/3.15576E7 (seconds per year) = 13.4 yearly average terawatts of thermal energy

BUT we have to calculate the **electric** energy generated using the thermodynamic efficiency factor of 0.34, thus:

(0.34)(13.4) = 4.56 yearly average terawatts of electricity from breeder reactor nuclear technology.

This number is the absolute maximum electric power that can ever be obtained by nuclear, **PERIOD**. Compare this to the **current** world demands for energy of 16.4 yearly average terawatts. This nuclear electric power is only 28% of the current energy demand. I do not call this a solution to our future problems. It will certainly be helpful but not the total solution.

The life of this nuclear supply is in its favor and is:

10E6 metric tons/5,800 metric tons/yr = 1,724 years (not too bad!)

This is good, but who cares when it would not even meet our current requirements. Sorry, but there is no pot of gold at the end of this rainbow. Breeder reactor nuclear technology will not save our bacon.

The size of each of the 1,000 fast neutron breeder reactors would have been 4.56 gigawatts of electricity.

Erie, PA, March 2013

Don Halcom has a Ph.D in Chemical Engineering. He is in his mid-70s, and retired. He is available to respond to any follow up questions you may have. You can reach him here: drdon (dot) halcom (at) verizon (dot) net

You can read three earlier essays of his here: The Re-Creation, The Return to Feudalism, What Makes You Think We Can Grow Out of This?