How Much 235U is Used in a 2700MW Fission Reactor? Answer in kg/day

[desibabu90]

A fission reactor operates at 2700 MW level. Assume all this energy comes from the 200 MeV released by fission caused by thermal neutron absorption by 235U. At what daily rate is the mass of 235U used? (In practice, of course, the energy conversion is not 100% efficient, nor is all the 235U in a fuel cell used.)

Answer is in kg/day

I really don't know how to solve this question. Am I supposed to convert the 2700MW first into MeV/day and then get it into kg by using Avagadro's Number? I have no idea. Please help!

If you are supposed to do that then u would do:
2700MW * 1e6 (convert into watts) * (1ev/(1.6e-19)) * (1MeV/1e6)*(86400s/day) ?

 

[mgb_phys]

Exactly, just convert 2700MW into J/s
Then work out how many 200Mev this is per second, then per day
Finally how many kg of U235 is this many atoms ( using Avagaro's number)

hint - it might help to have an order of magnitude estimate before you put so many large numbers into your calculator

 

[desibabu90]

what I don't understand is what the 200MeV energy that is created by fission is used for? is it suppose to equal something?

 

[mgb_phys]

what I don't understand is what the 200MeV energy that is created by fission is used for?

Making electricity - that's not really part of the question

 

[vanesch]

Moved: this is homework.

 

[desibabu90]

ok i am still not getting the answer...

So this is what I am doing...

http://img404.imageshack.us/img404/1023/webass.png

after that i don't know how to factor avagadro's number...so i just straight out multiply or do i need to use the molar mass. if i use the molar mass then the kgs cancel out...

plus am i using the conversion factor of 1.6605e-27 kg / 931.5Mev/c2 correctly?PS: sorry about it being in the wrong section..

 

[mgb_phys]

2700MW = 2.7 E9 J/s
200 Mev = 200E6 * 1.6E-19 J = 3.2E-11 J
This gives you the number of U235 fissions/second and so per day - that you need to make to generate that amount of power.

You are just using the 200Mev to count the number of U235 used up.
Each U235 atom has a mass of 235 amu, 1 kg is 6.02e23 amu (this is avagadros number)
Then it's just a case of being careful entering the numbers.

hint - how big an answer do you expect?

 

[desibabu90]

lol

"2700MW = 2.7 E9 J/s" ---- Got that!

"200 Mev = 200E6 * 1.6E-19 J = 3.2E-11 J" ------- Got that!

"This gives you the number of U235 fissions/second and so per day - that you need to make to generate that amount of power." --- what does -- dividing 2.7e9J/s by 3.2e-11?? i am really sorry I am being an idiot right now..i have lots of midterms next week and just can't focus properly

You are just using the 200Mev to count the number of U235 used up.

Each U235 atom has a mass of 235 amu, 1 kg is 6.02e23 amu (this is avagadros number) ---- isn't 1g = 6.02e23 amu so 1kg = 6.02e26? rite?

Then it's just a case of being careful entering the numbers.

hint - how big an answer do you expect? ---- I really don't know how big to expect -

 

[mgb_phys]

Yes, one 200Mev event is formed by a breakdown of a single U235 atom.
So to get 2.7E9 J/s from 3.2E-11J events you need 2.7E9/3.2E-11 of them per second
.
Correct - a mole is the number of atoms in 235grams of U235 (that's what I meant about checking numbers)
Finally you know that you don't need train loads of uranium arriving at a power station every day, so you can guess whether you expect an answer of a few grams or a few tons.

 

[desibabu90]

hey thanks a lot for ur help it worked

sorry for being an idiot

lol