13.18 A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much
did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of and that this nuclide is consumed only by the fission process.
13.18 A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much
did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of and that this nuclide is consumed only by the fission process.
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1 Answer
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13.18 Half life of the fuel in the fission reactor,
= 5 years = 5= 157.68
We know that in the fission of 1 g of
, the energy released = 200 MeV1 mole i.e. 235 gm of contains 6.023 atoms
Therefore 1 gm of
contains == 2.563 atomsThe total energy Q generated per gm of = 2002.563 MeV/g = 5.126
MeV/g = 5.126 J/g = 8.20
J/gSince the reactor operates only 80% of the time, hence the amount of
in 5 years is given by = 1538 kgHence, initial amount of fuel = 2 = 3076 kg
Similar Questions for you
Q = [4 *4.0026 – 15.9994] *931.5 MeV
Q = 10.2 MeV
-(1)
for B,
for B,
-(2)
The reaction is X²? → Y¹²? + Z¹²?
Binding energies per nucleon are: X=7.6 MeV, Y=8.5 MeV, Z=8.5 MeV.
Gain in binding energy (Q) = (Binding energy of products) - (Binding energy of reactants)
Q = (120 × 8.5 + 120 × 8.5) - (240 × 7.6) MeV
Q = (2 × 120 × 8.5) - (240 × 7.6) MeV = 2040 - 1824 = 216 MeV.
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