Is this nuclear physics question's answer correct?

[Toyona10]

Homework Statement

What is the power output of a reactor fueled by uranium-235 if it takes 30 days to use up 10 Kg of fuel and if each fission gives 200 MeV of energy.

Homework Equations

The Attempt at a Solution

(I was absent that day so this answer is from someone else's notebook)
No of U-235 nuclei in 10 kg:
(6.023*10^23*10000)/235 = 2.56*10^25

Fission rate= (2.56*10^25)/(30*24*3600)= 9.76*10^18 J/s (?? how is it joules per second, isn't that number above just no. of nuclei?)

Therefore, energy produced through fission per second:
= 200* 9.76*10^18= 1.95*10^15 MW

So...is this correct?
Thank you

 

[fzero]

Homework Statement

What is the power output of a reactor fueled by uranium-235 if it takes 30 days to use up 10 Kg of fuel and if each fission gives 200 MeV of energy.

Homework Equations

The Attempt at a Solution

(I was absent that day so this answer is from someone else's notebook)
No of U-235 nuclei in 10 kg:
(6.023*10^23*10000)/235 = 2.56*10^25

Fission rate= (2.56*10^25)/(30*24*3600)= 9.76*10^18 J/s (?? how is it joules per second, isn't that number above just no. of nuclei?)

Therefore, energy produced through fission per second:
= 200* 9.76*10^18= 1.95*10^15 MW

So...is this correct?
Thank you

As you notice above, because those calculations don't carry the units through from left to right, it's hard to figure out if things are correct by dimensional analysis alone. So you will want to start out by defining each factor introduced along with their correct units. When you do this, you quickly realize that the calculations don't make sense because the units don't work out.

For example,

(6.023*10^23*10000)/235 = 2.56*10^25

is missing a conversion from atomic mass units to grams. The fission rate is, as you noticed, not in J/s. Finally

= 200* 9.76*10^18= 1.95*10^15 MW

doesn't account for the fact that when we list an energy in MeV, we really mean MeV/c^2, where c is the speed of light. In order to obtain an expression in SI units, we need to use the value of c to convert units.

There could be other problems, but these were the most obvious ones. Hopefully it gives you a good idea of how to do the correct calculation.

 

[Toyona10]

As you notice above, because those calculations don't carry the units through from left to right, it's hard to figure out if things are correct by dimensional analysis alone. So you will want to start out by defining each factor introduced along with their correct units. When you do this, you quickly realize that the calculations don't make sense because the units don't work out.

For example,

(6.023*10^23*10000)/235 = 2.56*10^25

is missing a conversion from atomic mass units to grams. The fission rate is, as you noticed, not in J/s. Finally

= 200* 9.76*10^18= 1.95*10^15 MW

doesn't account for the fact that when we list an energy in MeV, we really mean MeV/c^2, where c is the speed of light. In order to obtain an expression in SI units, we need to use the value of c to convert units.

There could be other problems, but these were the most obvious ones. Hopefully it gives you a good idea of how to do the correct calculation.

Hmmm...lol

well, this is how I tried it:
total energy= energy per fission*no of atoms
= 200 MeV * 2.56*10^25
=8.19*10^14 J
so P_out=total E / time =(8.19*10^14)/(30*24*3600)
=315.97MW

 

[fzero]

Hmmm...lol

well, this is how I tried it:
total energy= energy per fission*no of atoms
= 200 MeV * 2.56*10^25
=8.19*10^14 J
so P_out=total E / time =(8.19*10^14)/(30*24*3600)
=315.97MW

That looks better, but I haven't looked up the precise conversion factors, so I'm just assuming you have them correct. I'd round it off to 2 or 3 significant digits, since the way the number of atoms was computed was rough.

 

[Toyona10]

agree

to which one? the 1st answer or the second one? :D