The Fort Calhoun nuclear power plant north of Omaha produces 476 MW of electrica

Question

The Fort Calhoun nuclear power plant north of Omaha produces 476 MW of electrical power from a 235U reactor core. The uranium isotope 235U produces 202.5 MeV of energy when it fissions.

a) If the plant is 33% efficient, how much power must the reactor produce?
b) How many atoms of 235U are consumed each second at Fort Calhoun?
c) What mass of 235U is consumed each second?
d) What mass of 235U is required for continuous production of power from this plant for a year?
e) The Sheldon power station in Hallam, NE produces 225 MW, less than half of that of Fort Calhoun. Look up how many tons of coal are used each year, and how many tons of CO2 would be released? (Sheldon power station uses 6 million tons of coal each year)

Explanation / Answer

a) electric power = 0.33 x reactor produced power

Power produced by reactor = 476 MW / 0.33 = 1442.42 MW

B) energy produced each second = 1442.42 x 10^6 J

energy produced by 1 atom = 202.5 x 10^6 x 1.6 x 10^-19 J

= 3.24 x 10^-11 J

atoms of 235U needed = 1442.42 x 10^6 / (3.24 x 10^-11 ) = 4.452 x 10^19 atoms

c)
there are 6.022 x 10^23 atoms in 1 mol

hence mol of 235U needed = 4.452 x 10^19 / (6.022 x 10^23) =7.393 x 10^-5 mol

mass of 1mol = 235 g

mass needed per second = 235 x 7.393 x 10^-5 = 0.0174 g

d) Seconds in a year = 365 x 24 x 3600 sec

mass needed = 0.0174 x 365 x 24 x 3600 = 547873.7 g Or 547.87 kg

e) 1 mol CO2 is produced per mol of carbon.

so 44 g CO2 is produced per 12 g C

hence tons of CO2 prod7ced = 6 million tons x 44 / 12 = 22 millions tons

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