Question

n this problem, we're going get a rough estimate the amount of uranium fuel it would take if the US recieved all its electrical power from nuclear power plants.

The size of a power plant in normally given as the about of electrical power it can produce when running a full capacity. This electrical power produced can be very different than the mechanical or thermal power that was required to produce this electricity. For example, power plant might have a \"thermal efficiency\" of 25\% and so require 100 MWt (mega-watts of thermal power) to produce 25MWe (megawatts of electrical power). The efficiency will vary from plant to plant but an approximate range is from around 2\5% to 35\%.

Lets assume we have a 10³ MWe electrical power plant that recieves its thermal energy from pressured water nuclear reactor (PWR) and has overall thermal efficiency of 30\%.

You may want to use the following table of atomic masses:

Table of masses
141Ba	140.9144 u		144Ba	143.9229 u		139Te	138.9347 u
141Cs	140.9196 u		90Kr	89.91952 u		91Kr	90.92344 u
92Kr	91.92615 u		94Zr	93.90632 u		93Rb	92.92157 u
235U	235.0439 u		p	1.00728 u		n	1.00867 u

A key point here is that not all of the uranium fuel in the reactor is ²³⁵U. Most of it is actually a different isotope, ²³⁸U, which does not fission in standard reactors. Lets assume the fuel is \"enriched\" so that 2.8\% of the fuel is actually ²³⁵U by mass. What is the total mass of fuel is used in one year?

Use_total =

Assume that all the fuel used in one year must actually be removed as high level radioactive waste. What volume of waste must be removed from the reactor annually and placed in long term storage?

V_total =

Take the electrical production of the US to be around 2.5X10¹²kWh/year. If all of the electrical power was generated by nuclear power plants similar to the one described above, what would the amount of waste that would need to be stored annually?

V_national =

If this waste were formed into a cube, what would be the length of the cube's sides?

L =

Answer 1

Let us first calculate the mass and power extracted for a single U235 nucleus.

A U235 nucleus contains 143 neutrons and 92 protons. When these number of nucleons combine to form a single U235 nucleus some of the mass get lost to give binding energy of the nucleus.

The mass loss = 135.0439 – (143 x 1.00867 + 92 x 1.00728) = 1.86567 u

Binding energy = mass loss x c² = 1.492418775 x 10^-10 J

The power generated by one U235 = 4.145607708 x 10^-14 Wh. (As 1 Wh energy is equivalent to 3600 J)

Or inversely to generate 1 Wh electrical energy one needs ~ 2.412192 x 10¹³ U235 fuel atoms.

So

1 Wh = 2.412192 x 10¹³ U235 fuel atoms (approximately).

In a year US needs 2.5 x 1012 x 10³ Wh electrical energy. (If I read it right)

So it requires 2.5 x 1012 x 10³ x 2.412192 x 10¹³ U235 fuel atoms/year

= 6.10284576 x 10¹⁹ U235 fuel atoms = 1.013425 x 10^-4 moles of U235 /year (N_A = 6.022 x 10²³)

Density of U235 = 235.044 gm / mol

So the mass of required fuel = 2.381994657 x 10^-2 gm.

But as only 2.8 % of the fuel is U235 so total mass of fuel = 2.381994657 x 10^-2 / 2.8 % gm

Usetotal = 0.8507 gm

The density of U235 is 19.1 gm/cm³

This gives the volume of the fuel used

Vtotal = 0.04454 cm³.

If this is formed into a cube the length of the cube will be

L = 0.35447 cm