n this problem, we\'re going get a rough estimate the amount of uranium fuel it

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% 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

1. What is the mass of 235U fissioned in one year?

Use =

2. 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 2.9% of the fuel is actually 235U by mass. What is the total mass of fuel is used in one year?

Usetotal =

3. 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?

Vtotal =

4. 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?

Vnational =

5.

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

L =

Explanation / Answer

1) we have 103MWe power as electrical output from 30% thermal efficiency.

That means the input energy =103-103*30/100 =72.1MWt

now 1u = Mu/Na   , 103MWe= 103*86.4*10^9J/d , therefore , 103MWe= 103*86.4*10^9* 365= 3.2*10^15J/y

considering neutron bombarding on 235U we get 141Ba,92Kr, 3n therefore,

energy per fission,

235.0439u- [ 140.9144+ 91.92615+ 3*1.00807]u = .82031u

now 1u= 1.6604 810^-27kg, .82031u =.82031* 1.6604 8*10^-27kg

using E=mc^2 we get .82031u= 1.2258 *10^-10J

therefore,

we get 1.2258 *10^-10J energy from .82031* 1.6604 8*10^-27kg 235U

weget 1J energy from .82031* 1.6604 8*10^-27/1.2258 *10^-10 kg

weget 3.2* 10^15J   energy from .82031* 1.6604 8*10^-27/1.2258 *10^-10 *3.2*10^15Kg=3.5*10^-2Kg

2) if 2.9% of atomic mass of235U is used that means the actual mass per fission becomes,

.82031* 2.9/100 = .0237* 1.664 *10^-27 Kg

for 1 year total used fuel=.0237*1.664*10^-27/.0237 *1.4944*10^-10*3.2*10^15 Kg= 3.56*10^-2Kg

3) considering high level radioaactive waste and used input mass per year 3.56*10^-2Kg

atomic density of 238U is W238* *Na /M238 = (100-2.9)*19.1*6.022*10^23/238.0508 =4.69*10^23

Therefore total volume= M/D= 3.56*10^-2/4.69*10^23 cm^3=.75*10^-23 cm^3

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