Calculating amount of fissionable material needed for reactor

In summary, using the equations Power = Energy/Time and E=mc^2, the energy produced by a 300 MW fission reactor is calculated to be 1.62 x 10^23 MeV, resulting in a mass of 0.250 Kg of 235U being consumed in one day. However, accounting for the 30% efficiency of the reactor, the actual mass consumed is 834.98 Kg.
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leviathanX777
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Homework Statement


A fission reactor produces 300 MW of electrical power with an efficiency of 30%. Calculate the mass of 235U consumed by the reactor in one day.


Homework Equations


Power = Energy/Time

E=mc^2

The Attempt at a Solution


Using the above equations I got an energy of 1.62 x 10^23 MeV. The corresponding mass found using the fact that 1 amu = 1.66 X 10^-27 Kg was 0.250 Kg. Is this right? I don't see how to bring the efficiency of 30% into this. Is it basically 30% of the energy that I calculated?
 
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  • #2
Ah wait, I realized that I needed to add another 70% onto the 300 MW which gave me 1,000 MW. However the answer I got was 834.98 Kg, seems pretty high to be correct. Am I doing something wrong?
 

1. How is the amount of fissionable material needed for a reactor calculated?

The amount of fissionable material needed for a reactor is calculated based on the desired power output of the reactor, the type of fissionable material being used, and the efficiency of the reactor. The formula for calculating the amount of material needed is: Amount of material (in kilograms) = Power output (in megawatts) / Efficiency (in decimal form) * Fission energy (in joules per kilogram).

2. What is the most commonly used fissionable material in nuclear reactors?

The most commonly used fissionable material in nuclear reactors is uranium-235. This is because it is relatively abundant and has a high rate of fission when bombarded with neutrons.

3. How does the efficiency of a reactor impact the amount of fissionable material needed?

The efficiency of a reactor directly impacts the amount of fissionable material needed. A more efficient reactor will require less material to produce the same amount of power, while a less efficient reactor will require more material. This is why it is important for scientists to continually improve reactor efficiency.

4. Are there any safety concerns when calculating the amount of fissionable material needed for a reactor?

Yes, safety concerns must be taken into account when calculating the amount of fissionable material needed for a reactor. Too much material can lead to a runaway reaction, causing a nuclear meltdown, while too little material can result in a reactor that is unable to sustain a chain reaction.

5. Can the amount of fissionable material needed for a reactor change over time?

Yes, the amount of fissionable material needed for a reactor can change over time. This can be due to advancements in technology that increase reactor efficiency, changes in power output requirements, or the availability of different types of fissionable material. Scientists must continually reassess and adjust the amount of material needed for reactors to ensure safe and efficient operation.

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