Honors Nuclear Reactions

Nearly all of you do not understand at least some fundamental part of the relations between quantities.

Print this out and study it.


If you put units in your equations and do dimensional analysis you will avoid some of the more inane mistakes you are currently making. Dimensional analysis means when you do the algebra on the units the result should yield the units you need for the quantity you are finding.
(m3 means cubic meter here) ( per sec is the same as /sec, etc.)
1) Power is energy transferred or transformed per unit of time or J/s
2) If you know the reactions per sec per m3 and you know the energy per reaction you should know how to find the energy per sec per m3. Think about it and make sure you can do it.
3) The energy per m3 per sec is the power per m3, i.e. J/s per m3 = W/m3. By the way, given the amount of fuel per m3 and the sigmav Whcih by the way has units m3/sec) the reactions per m3 have nothing to do with the volume of the plant. The total number of reactions in the plant of course = the reactions per m3 time the volume. You must think and understand rather than try to remember a jumble of formulae or how we did old problems.
4) If 1 MW is one million W than you must know that you must divide the number of W by 1 million to get the number of MW. After all, don't you divide the number of cents by 100 to get the number of dollars, or is this news to you?
5) The power output of a plant is the fusion ( or other microscopic) energy generated by the plant per sec times the efficiency. The fusion power generated by a fusion plant is of course the fusion power per cubic m times the volume of the reactor. Thus output = power per m3 x volume times efficiency
6) The mass of fuel in anything = number of particles per m3 x mass per particle times volume of the thing.
7)The kg of fusion fuel used per day is reactions per sec per m3 x volume x sec per day x the mass of the fuel particles per each reaction ( 5 x mp for DT)
8)To find kg of coal used per day, given J/kg of coal clearly you need to find the Joules needed per day and to do that you can take the output power ( W = J/s) times sec per day and then you figure out how to find out how much fuel you use because I won't tell you how to do things you should have learned 4 years ago. I will tell you that by now you should be able to convert MW to J/s. Aso, dimesional analysis may help if you do it right.
9) You must get some understanding of what numbers mean. You are not to write down whatever nonesense you get out of your calculator. A power plant does not use a mass of fuel greater that the mass of the entire galaxy. 10 to the 37th power kg is not correct! In the future answers displaying this level of not caring will result in loss of credit for the entire problem.