Calculating Time to Reach 100 kW Output from Uranium-235 Fission

In summary, to go from the first fission to 100 Kw output, 31250000000000 reactions per second are needed. Using logarithms, the number of generations needed can be found to a fraction, but without using logarithms, starting with the initial number of reactions per second and doubling until reaching the desired rate will give the nearest whole number of generations, which can then be used to calculate the time required.
  • #1
whiteshado
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During the fission of an atom of uranium-235, 200 Mev are released. If the average time between generations of fission reaction is 10e-14 seconds and if each fission reaction gives rise to two fissions in the next generation, calculate the time required to go from the first fission to 100 Kw output.

200mev=200000000ev=0.000,000,000,032jouls per sec
1kw=1000joules
31250000000000reactions per second to make 1kw
now my thinking was (long version) 1x2 till i reached 3125000000000000 and took how many x2s it took to reach 3125000000000000 and x taht by 10e-14 but taht didnt work any help is appreaciated
i know its easy (or it should be) i just can't figure it out at the moment
 
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  • #2
perhaps this question is harder than i think for the great minds that roam these fourms cannot direct me in the correct direction
 
  • #3
i formula perhaps? maybe a hint like what number it starts with? anything? no? sigh...
 
  • #4
wow no one could help me with this question?
 
  • #5
whiteshado said:
During the fission of an atom of uranium-235, 200 Mev are released. If the average time between generations of fission reaction is 10e-14 seconds and if each fission reaction gives rise to two fissions in the next generation, calculate the time required to go from the first fission to 100 Kw output.

200mev=200000000ev=0.000,000,000,032jouls per sec
1kw=1000joules
31250000000000reactions per second to make 1kw
now my thinking was (long version) 1x2 till i reached 3125000000000000 and took how many x2s it took to reach 3125000000000000 and x taht by 10e-14 but taht didnt work any help is appreaciated
i know its easy (or it should be) i just can't figure it out at the moment

All those zeros are too scary for most of us to deal with. Your approach is basically correct, but it would be easy to miscount the number of generations. How many x2s do you think it takes?

Do you know about logarithms and exponentials?
 
  • #6
yes i know about exponentials (notation correct?) umm i think i got to 45 or something this was yesterday
logarithms i might know about them
3.13e13=1kw 3.13e15=100kw
i can change all the zeros to 3.13e15 if you like and 3.2e-11 there you go
 
  • #7
whiteshado said:
yes i know about exponentials (notation correct?) umm i think i got to 45 or something this was yesterday
logarithms i might know about them
3.13e13=1kw 3.13e15=100kw
i can change all the zeros to 3.13e15 if you like and 3.2e-11 there you go

Ideally, you would like to use logarithms for this problem, but your approach would work to get the answer to the nearest whole number of generations. 45 generations is way more than the number needed. I agree with your calculation of the number of reactions per second that you need to achieve. To get the number of generations, without using logarithms, you need to start with the initial number of reactions per second and double that until the rate is high enough. You need to check your calculation of the original rate (reactions per second) and the number of doubling generations. If you use logarithms, you can find the number of generations to a fraction instead of a whole number. If you don't know how to do that, and want to, we can return to it after you get the nearest (or better yet, rounded up) whole number correct.
 

Related to Calculating Time to Reach 100 kW Output from Uranium-235 Fission

1. How is the time to reach 100 kW output calculated?

The time to reach 100 kW output is calculated by using the formula: t = (ln(100/1) / λ), where t is the time in seconds and λ is the decay constant of Uranium-235, which is approximately 0.693/yr.

2. What is the significance of reaching 100 kW output from Uranium-235 fission?

Reaching 100 kW output from Uranium-235 fission is significant because it is the minimum amount of power needed to sustain a nuclear reactor and generate electricity. This is known as the critical point.

3. How long does it typically take to reach 100 kW output from Uranium-235 fission?

The time to reach 100 kW output from Uranium-235 fission depends on the size and efficiency of the reactor, but it usually takes a few hours to reach this level of output.

4. Are there any factors that can affect the time to reach 100 kW output?

Yes, there are several factors that can affect the time to reach 100 kW output from Uranium-235 fission, such as the purity and enrichment level of the uranium fuel, the design and size of the reactor, and the control mechanisms used to regulate the fission process.

5. Can the time to reach 100 kW output be shortened or prolonged?

Yes, the time to reach 100 kW output can be shortened by increasing the enrichment level of the uranium fuel or by using more efficient reactor designs. It can also be prolonged by using lower enriched fuel or less efficient reactor designs. Additionally, external factors such as maintenance or accidents can also affect the time to reach 100 kW output.

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