UFOEEK
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  1. Attachment 655378The question is when a Uranium-235 nucleus undergoes fission ,approximetly 200 MeVof energy is released . Estimate the total mass of original fuel required per year in a 1600 MW nuclear reactor that uses enriched fuel containing 3% uranium -235 and 97% uranium-238 and operates at an efficiency of 25% .

    The mark scheme is a shambles so i really need help in understanding what on earth i have to do please
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Uni12345678
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what have you done so far ?
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UFOEEK
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(Original post by Uni12345678)
what have you done so far ?
ive only changed the 1600 MW to Joules but the mark scheme says that the 1600 MW is what is does at 100% efficiency then proceeds to multiply ot by 4 and call the answer 25% efficiency ? or ive completely misunderstood it :/ either way im stumped
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Uni12345678
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(Original post by UFOEEK)
ive only changed the 1600 MW to Joules but the mark scheme says that the 1600 MW is what is does at 100% efficiency then proceeds to multiply ot by 4 and call the answer 25% efficiency ? or ive completely misunderstood it :/ either way im stumped
hmmm okay, I divided by 4...
so that gives 400 MJ... then, I got 1.26 x 10^6 J released in one year
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PawanAviator
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(Original post by UFOEEK)
  1. The question is when a Uranium-235 nucleus undergoes fission ,approximetly 200 MeVof energy is released . Estimate the total mass of original fuel required per year in a 1600 MW nuclear reactor that uses enriched fuel containing 3% uranium -235 and 97% uranium-238 and operates at an efficiency of 25% .

    The mark scheme is a shambles so i really need help in understanding what on earth i have to do please
What's the answer??
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Dragolien
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I'm not entirely sure if this is how you're meant to go about this, but I'll try.

Converting the 200MeV to joules gives us 3.2043546e-11J. So dividing 1600000000J by that gives us about 5*10^19. This is the number of uranium-238 nuclei required per second. To convert this to years, multiply by 60*60*24*365 to get 1.6*10^27.

Now to convert this from 3% to 100%. Divide by 3 and multiply by 100 to get 5.25*10^28. And because the efficiency is only 25%, multiply by a final 4 to get 2.1*10^29 atoms of uranium.

Divide this by Avogadro's constant to get the number of moles. This gives you 348637.4 moles of uranium. Divide by the relative atomic mass to get 1464g, or 1.464kg per year.
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Uni12345678
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(Original post by UFOEEK)
  1. The question is when a Uranium-235 nucleus undergoes fission ,approximetly 200 MeVof energy is released . Estimate the total mass of original fuel required per year in a 1600 MW nuclear reactor that uses enriched fuel containing 3% uranium -235 and 97% uranium-238 and operates at an efficiency of 25% .

    The mark scheme is a shambles so i really need help in understanding what on earth i have to do please
could you post a photo of the mark scheme?
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UFOEEK
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Name:  Screenshot_20170531-195352.png
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Size:  428.0 KB its question b)
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UFOEEK
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thanks for the replies btw i appreciate any response 😥
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localwhiteasian
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what paper is this?
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UFOEEK
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(Original post by localwhiteasian)
what paper is this?
its from the questions by topic from Physics and Maths tutor .com
im assuming theyre all past paper qus from different specs in some cases ?
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UFOEEK
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(Original post by Uni12345678)
could you post a photo of the mark scheme?
i think i already did but ill do so againName:  Screenshot_20170531-195352.png
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Uni12345678
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(Original post by UFOEEK)
i think i already did but ill do so againName:  Screenshot_20170531-195352.png
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Oh you multiply by 4 because 1600 MW is the output power not the input
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UFOEEK
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(Original post by Uni12345678)
Oh you multiply by 4 because 1600 MW is the output power not the input
okay that makes sense i suppose its just funny wording
but then what
i kind of get what theyve done with the rest of it but i dont get how they got the energy from 1 kilo of fuel
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PawanAviator
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(Original post by UFOEEK)
  1. Attachment 655378The question is when a Uranium-235 nucleus undergoes fission ,approximetly 200 MeVof energy is released . Estimate the total mass of original fuel required per year in a 1600 MW nuclear reactor that uses enriched fuel containing 3% uranium -235 and 97% uranium-238 and operates at an efficiency of 25% .

    The mark scheme is a shambles so i really need help in understanding what on earth i have to do please
So let me show you how I did it. I done it slightly different to the mark scheme

So we know that 1 U-235 Nucleus releases 200MeV. I converted this into Joules

200 MeV = 200 x 1.6 x 10^-13 = 3.2 x 10^-11 J <= 1 U-235 nucleus

So to produce 1600 MW for 1 year ==> P = 1600 x 10^6 W and t = 365 x 24 x 60 x60

using E = P x t

E = (1600 x 10^6) x (365 x 24 x 60 x 60) = 5.04576 x 10^16 J of energy required

so we need 5.04576 x 10^16 / 3.2 x 10^-11 = 1.5768 x 10^27 U-235 nuclei
to produce 1600MW in 1 year


so now we can calculate the mass of this amount by calculating the number of moles using avogadro's constant.

1.5768 x 10^27 / 6.02 x 10^23 = 2619 moles of U-235

1 mole of U-235 is 0.235kg

there 2619 moles = 2619 x 0.235 = 615.5kg of U-235

but since the fuel rods contain 3% of this U-235 then the total amount = 615.5 / 0.03 = 20520 kg

Now since its 25% effcient then we need 4 times this amount to achieve this output

20520 x 4 = 8 x 10^4
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UFOEEK
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(Original post by PawanAviator)
So let me show you how I did it. I done it slightly different to the mark scheme

So we know that 1 U-235 Nucleus releases 200MeV. I converted this into Joules

200 MeV = 200 x 1.6 x 10^-13 = 3.2 x 10^-11 J <= 1 U-235 nucleus

So to produce 1600 MW for 1 year ==> P = 1600 x 10^6 W and t = 365 x 24 x 60 x60

using E = P x t

E = (1600 x 10^6) x (365 x 24 x 60 x 60) = 5.04576 x 10^16 J of energy required

so we need 5.04576 x 10^16 / 3.2 x 10^-11 = 1.5768 x 10^27 U-235 nuclei
to produce 1600MW in 1 year


so now we can calculate the mass of this amount by calculating the number of moles using avogadro's constant.

1.5768 x 10^27 / 6.02 x 10^23 = 2619 moles of U-235

1 mole of U-235 is 0.235kg

there 2619 moles = 2619 x 0.235 = 615.5kg of U-235

but since the fuel rods contain 3% of this U-235 then the total amount = 615.5 / 0.03 = 20520 kg

Now since its 25% effcient then we need 4 times this amount to achieve this output

20520 x 4 = 8 x 10^4
omdz thanks a bunch this makes so much sense!:dancing2::dancing2:
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Eimmanuel
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(Original post by UFOEEK)
i think i already did but ill do so againName:  Screenshot_20170531-195352.png
Views: 247
Size:  428.0 KB
If you are still interested in the what the MS is doing, you can continue to read.

200 × 1.6 × 10-13 is converting 200 MeV to the unit of joules.

(200 × 106 eV) × (1.6 × 10-19 J/eV) is simplified to 200 × 1.6 × 10-13

1/0.238 × 0.03 is to find out the number of moles Uranium-235 in 1 kg of fuel. The need of times 0.03 is because you are told the following:

...enriched fuel containing 3% uranium -235 and 97% uranium-238 ....

Thus in 1 kg fuel, only 0.03 kg is uranium -235.

1/0.238 × 0.03 × 6.0 × 1023 is to find out how many uranium -235 nuclei in 1 kg of the fuel.

Hope it helps.
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