The most promising fusion reaction for electrical power generation is the deuter

Question

The most promising fusion reaction for electrical power generation is the deuterium (2H)-tritium (3H) or D-7. reaction. hydrogen that would have to be artificially manufactured, but that proce simple ns needed for D-Tfusion Tritium is a radioactive form of is technologically energy carried a. Find Q for this reaction. released energy. The is much more b. The alpha carries about 20% and the neutron 80% of the neutron's energy by the particle charged can be captured the alpha (sinc it is complicated. Calculate the velocity of the neutron. c. One scheme for capturing the neutron of liquid lithium. The following reaction would occur: The alpha could then be easily captured and the iH used as fuel for additional reactions. Find the Q d. for this neutron capture reaction. fusion per year to Assuming all of this can be accomplished, how many kg of hydrogen must undergo supply a typical 20 MW power plant? Assume about 30% of total energy released by fusion and he neutron capture is converted to electricity. Mathematical Analysis

Explanation / Answer

a) exact mass of 2H and 3H are respectively 2.014 amu and 3.016amu

again exact mass of 4He and n are 4.002amu and 1.008amu

sum of the masses of reactants=2.014+3.016amu=5.03amu

sum of the masses of products=4.002+1.008amu=5.01amu

therefore change in mass=mass of product-mass of reactants=5.01-5.03 amu=-.02amu

so energy is released in this case then Q will be positive

therefore Q=.02amu*931.5Mev/amu=18.63Mev

b) neuton takes 80% of the released energy

therefore neutron energy=18.63*80/100=14.904Mev=14.90* 1.602*10^-13J

using the energy velocity relation

velocity of the neutron v=sqrt(2E/m)=sqrt(2* 14.90* 1.602*10^-13/1.675*10^-27)m/s=5.33*10^7m/s

neutron mass=1.675*10^-27kg

c)

mass of neutron and 6Li are 1.008amu and 6.011amu

sum of reactants=1.008+6.011amu=7.019amu

mass of 4He and 3H are 4.002 and 3.016amu

sum of products= 7.018amu

change in mass=7.018-7.019=-.001amu

again energy is released so Q is positive

Q=.001*931.5Mev=.9315Mev

d) from the first solution of a) total 5.03amu hydrogen releases 18.63Mev energy

means 5.03*1.66*10^-27kg hydoren releases 18.63Mev=18.63*4.450*10^-23MWh

now yearly output of 20MW power plant is 20*30/100*365*24=52560MWh

so,

18.63*4.450*10^-23MWh is released from 5.03*1.66*10^-27kg hydoren

or 52560MWh is released from (5.03*1.66*10^-27/18.63*4.450*10^-23)* 52560 kg hydrogen

=.5292Kg

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