Use the blog if you can understand it, but the fact that this blog can't show exponents or subscripts makes it hard. Here is a link to the same information in a Word document.
DHe3problem.doc
which should be easy to follow.
DHe3problem.doc
which should be easy to follow.
Follow the process, and make sure you understand it, don’t just fill in the blanks blindly. There will be a quiz on this early this week.
The deuterium - helium 3 (D He3) reaction is perhaps the most desirable of all the easily achievable ones. The products are helium 4 and a proton. These particles are charged and can be confined by magnetic fields. They also do not interact with nuclei in the materials around them ( i.e., the reactor vessel) to make them radioactive.
Let’s investigate how many reactions per second and then how many fuel nuclei per m3 it would take for a D He 3 fueled power plant to put out 1 billion watts. Assume 40% efficiency so we need
10E9(J /sec )/.4 = ____ joules /sec of energy from the reactions.
Each reaction produces 18.4 MeV or ______J
The number of reactions per sec x the energy per reaction = energy per sec from the reactions.
Therefore the number of reactions we need per second is found from
________/_________ = _________ reactions per second
This reactor will have to be substantially larger than the deuterium tritium reactor we talked about Friday; lets say 250 m3 instead of 100 m3. The number of reactions we will need per m3 per sec is simply the number of reactions per sec / volume of reactor. So in this case the required reaction rate = _________ reactions per sec per m3.
Remembering that the reaction rate i.e. the number of reactions per m3 per sec is given by the number of “bullets” times their speed times the number of targets x the area per target, i.e. we have:
reaction rate [reactions/(m3s)] = nD vD nHe3 sigma; where n is the number of nuclei per m3 (number density), vD is the average speed of the deuterium nuclei and sigma is the target area per nucleus. Recall that for maximum reaction rate we set the two “n”s equal to each other and therefore each n = 1/2 the total fuel density or nD = nHe3= nfuel/2.
Lets look at the velocity. With the DHe3 reaction we have one nucleon with one proton and one neutron, the D: and one nucleon with two protons and one neutron, the He3. Therefore the potential energy when the two fuel nuclei approach each other is now given by PE = kcQ1Q2/r [kc = 9 billion or 9E9]
9E9x1.6E-19 x 2x1.6E-19/ r . If we approximate r as 2.E-14 [ this may be a poor estimate and we will discuss the details of this tomorrow] we get PE = ______J.
This PE of the nuclei as they reach this close distance, must be equal to the KE they started with, so we can use this KE and the fact that mD x vD2/2 = KE to find vD of the deuterons circulating in the reactor. Recall that the mass of a deuteron, which contains two nucleons, is 2 x 1.67E-27, and rewrite the definition for KE to find vD and get vD =____________m/s.
From the literature we find sigma is about 7E-29 m2.
Rewriting the reaction rate equation as (nfuel squared/4) x vD x sigma and then rewriting it again so we can find nfuel ,
we have nfuel2/4= required reaction rate [i.e. reactions per m3per sec]/( vD x sigma) Multiplying both sides by 4 and taking the square root of both sides we find nfuel = ______ fuel nucleons per m3.
Congratulations, you just completed your second fusion fuel density calculation, and two weeks ago you couldn’t even define fusion.
Let’s investigate how many reactions per second and then how many fuel nuclei per m3 it would take for a D He 3 fueled power plant to put out 1 billion watts. Assume 40% efficiency so we need
10E9(J /sec )/.4 = ____ joules /sec of energy from the reactions.
Each reaction produces 18.4 MeV or ______J
The number of reactions per sec x the energy per reaction = energy per sec from the reactions.
Therefore the number of reactions we need per second is found from
________/_________ = _________ reactions per second
This reactor will have to be substantially larger than the deuterium tritium reactor we talked about Friday; lets say 250 m3 instead of 100 m3. The number of reactions we will need per m3 per sec is simply the number of reactions per sec / volume of reactor. So in this case the required reaction rate = _________ reactions per sec per m3.
Remembering that the reaction rate i.e. the number of reactions per m3 per sec is given by the number of “bullets” times their speed times the number of targets x the area per target, i.e. we have:
reaction rate [reactions/(m3s)] = nD vD nHe3 sigma; where n is the number of nuclei per m3 (number density), vD is the average speed of the deuterium nuclei and sigma is the target area per nucleus. Recall that for maximum reaction rate we set the two “n”s equal to each other and therefore each n = 1/2 the total fuel density or nD = nHe3= nfuel/2.
Lets look at the velocity. With the DHe3 reaction we have one nucleon with one proton and one neutron, the D: and one nucleon with two protons and one neutron, the He3. Therefore the potential energy when the two fuel nuclei approach each other is now given by PE = kcQ1Q2/r [kc = 9 billion or 9E9]
9E9x1.6E-19 x 2x1.6E-19/ r . If we approximate r as 2.E-14 [ this may be a poor estimate and we will discuss the details of this tomorrow] we get PE = ______J.
This PE of the nuclei as they reach this close distance, must be equal to the KE they started with, so we can use this KE and the fact that mD x vD2/2 = KE to find vD of the deuterons circulating in the reactor. Recall that the mass of a deuteron, which contains two nucleons, is 2 x 1.67E-27, and rewrite the definition for KE to find vD and get vD =____________m/s.
From the literature we find sigma is about 7E-29 m2.
Rewriting the reaction rate equation as (nfuel squared/4) x vD x sigma and then rewriting it again so we can find nfuel ,
we have nfuel2/4= required reaction rate [i.e. reactions per m3per sec]/( vD x sigma) Multiplying both sides by 4 and taking the square root of both sides we find nfuel = ______ fuel nucleons per m3.
Congratulations, you just completed your second fusion fuel density calculation, and two weeks ago you couldn’t even define fusion.
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