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v\\:* {behavior:url(#default#VML);}o\\:* {behavior:url(#default#VML);}w\\:* {behavior:url(#default#VML);}.shape {behavior:url(#default#VML);}Normal0falsefalsefalseEN-USX-NONEX-NONEMicrosoftInternetExplorer4/* Style Definitions */table.MsoNormalTable{mso-style-name:Table Normal;mso-tstyle-rowband-size:0;mso-tstyle-colband-size:0;mso-style-noshow:yes;mso-style-priority:99;mso-style-qformat:yes;mso-style-parent:;mso-padding-alt:0in 5.4pt 0in 5.4pt;mso-para-margin-top:0in;mso-para-margin-right:0in;mso-para-margin-bottom:10.0pt;mso-para-margin-left:0in;line-height:115%;mso-pagination:widow-orphan;font-size:11.0pt;font-family:Calibri,sans-serif;mso-ascii-font-family:Calibri;mso-ascii-theme-font:minor-latin;mso-fareast-font-family:Times New Roman;mso-fareast-theme-font:minor-fareast;mso-hansi-font-family:Calibri;mso-hansi-theme-font:minor-latin;}It is proposed to use nuclear fusion reaction;2H1 + 2H1------->4He2in a nuclear reactor 200 MW rating. If the energy from the above reaction is used with a 25 % efficiency in the reactor, how many grams of deuterium fuel will be needed per day? ( The masses of 2H1 and 4He2 are 2.0141 atomic mass units and 4.0026 atomic mass units respectively)

Dhruv Gupta , 16 Years ago
Grade 12
anser 1 Answers
Gaurav Sharma

Last Activity: 16 Years ago

Mass defect in the given nuclear reaction:

 

                Dm = 2(mass of deuterium) – (mass of helium)

 

                        = 2(2.0141) – (4.0026)

 

                        = 0.0256

 

        Therefore, energy released

 

                DE = (Dm) (931.48) MeV = 23.85 MeV

 

                = 23.85 × 1.6 × 10–13 J

 

                = 3.82 × 10–12 J

 

        Efficiency is only 25%, therefore,

 

                25% of ?E = (25/100) (3.82 × 10–12) J = 9.55 × 10–13 J

 

i.e., by the fusion of two deuterium nuclei, 9.55 × 10–13 J energy is available to the nuclear reactor.

 

Total energy required in one day to run the reactor with a given power of 200 MW:

 

                ETotal = 200 × 106 × 24 × 3600 = 1.728 × 1013 J

 

        \ Total number of deuterium nuclei required for this purpose:

 

                n = Etotal/?E/2 = 2 x 1.728 x 1013 / 9.55 x 1013= 0.326 × 1026

 

        \ Mass of deuterium required

 

                = (Number of gm-moles of deuterium required) × 2 g

 

                = (0.362 x 1016)/(6.02 x 1023)× 2 = 120.26 g.

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