13.31 Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of ${}^{235}U$ to be about 200MeV.

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Payal Gupta | Contributor-Level 10

5 months ago

13.31 Amount of Electric power to be generated, P = 200000 MW

10% of which to be obtained from nuclear power.

Hence, amount of nuclear power = 10% of 2 $* 10^{5}$ MW = 2 $* 10^{4}$ MW = 2 $* 10^{4} * 10^{6}$ J/s

= 2 $* 10^{4} * 10^{6} * 60 * 60 * 24 * 365$ J/y = 6.3072 $* 10^{17}$ J/y

Heat energy release per fission of a ${}^{235}U$ nucleus, E = 200 MeV

Efficiency of a reactor = 25%

So the amount of electrical energy converted from heat energy per fission = 25% of 200 MeV = 50 MeV

= 50 $* 10^{6}$ eV = 50 $* 10^{6} * 1.6 * 10^{- 19}$ J = 8 $* 10^{- 12}$ J

Therefore, number of atoms required per year = $\frac{6.3072 * 10^{17}}{8 * 10^{- 12}}$ = 7.884 $* 10^{28}$

1 mole of ${}^{235}U$ = 235 gm of ${}^{235}U$ contains 6.023 $* 10^{23}$ atoms

Hence the mass of 7.884 $* 10^{28} a t o m s$ = $\frac{235 * 10^{-3}}{6.023 * 10^{23}}$ $*$ 7.884 $* 10^{28}$ = 30.76 $* 10^{3}$ kg

Hence, the Urani

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13.31 Amount of Electric power to be generated, P = 200000 MW

10% of which to be obtained from nuclear power.

Hence, amount of nuclear power = 10% of 2 $* 10^{5}$ MW = 2 $* 10^{4}$ MW = 2 $* 10^{4} * 10^{6}$ J/s

= 2 $* 10^{4} * 10^{6} * 60 * 60 * 24 * 365$ J/y = 6.3072 $* 10^{17}$ J/y

Heat energy release per fission of a ${}^{235}U$ nucleus, E = 200 MeV

Efficiency of a reactor = 25%

So the amount of electrical energy converted from heat energy per fission = 25% of 200 MeV = 50 MeV

= 50 $* 10^{6}$ eV = 50 $* 10^{6} * 1.6 * 10^{- 19}$ J = 8 $* 10^{- 12}$ J

Therefore, number of atoms required per year = $\frac{6.3072 * 10^{17}}{8 * 10^{- 12}}$ = 7.884 $* 10^{28}$

1 mole of ${}^{235}U$ = 235 gm of ${}^{235}U$ contains 6.023 $* 10^{23}$ atoms

Hence the mass of 7.884 $* 10^{28} a t o m s$ = $\frac{235 * 10^{-3}}{6.023 * 10^{23}}$ $*$ 7.884 $* 10^{28}$ = 30.76 $* 10^{3}$ kg

Hence, the Uranium needed per year is 30.76 $* 10^{3}$ kg

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13.31 Amount of Electric power to be generated, P = 200000 MW10% of which to be obtained from nuclear power.Hence, amount of nuclear power = 10% of 2 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>5</mn> </mrow> </mrow> </msup> </math>  MW = 2 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </msup> </math> MW = 2 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </msup> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </msup> </math>  J/s= 2 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </msup> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>60</mn> <mo>×</mo> <mn>60</mn> <mo>×</mo> <mn>24</mn> <mo>×</mo> <mn>365</mn> </math>  J/y = 6.3072 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>17</mn> </mrow> </mrow> </msup> </math> J/yHeat energy release per fission of a <math> <mmultiscripts> <mrow> <mrow> <mi>U</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow></mrow> </mrow> <mrow> <mrow> <mn>235</mn> </mrow> </mrow> </mmultiscripts> </math>  nucleus, E = 200 MeVEfficiency of a reactor = 25%So the amount of electrical energy converted from heat energy per fission = 25% of 200 MeV = 50 MeV= 50 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </msup> </math> eV = 50 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>1.6</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>19</mn> </mrow> </mrow> </msup> </math> J = 8 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>12</mn> </mrow> </mrow> </msup> </math>  JTherefore, number of atoms required per year = <math> <mfrac> <mrow> <mrow> <mn>6.3072</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>17</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>8</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>12</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math>  = 7.884 <math> <mo>×</mo> <mi> </mi> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>28</mn> </mrow> </mrow> </msup> </math> 1 mole of <math> <mmultiscripts> <mrow> <mrow> <mi>U</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow></mrow> </mrow> <mrow> <mrow> <mn>235</mn> </mrow> </mrow> </mmultiscripts> </math>  = 235 gm of <math> <mmultiscripts> <mrow> <mrow> <mi>U</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow></mrow> </mrow> <mrow> <mrow> <mn>235</mn> </mrow> </mrow> </mmultiscripts> </math>  contains 6.023 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>23</mn> </mrow> </mrow> </msup> </math> atomsHence the mass of 7.884 <math> <mo>×</mo> <mi> </mi> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>28</mn> </mrow> </mrow> </msup> <mi> </mi> <mi>a</mi> <mi>t</mi> <mi>o</mi> <mi>m</mi> <mi>s</mi> </math>  =<math><mfrac><mrow><mrow><mn>235</mn><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>-3</mn></mrow></mrow></msup></mrow></mrow><mrow><mrow><mn>6.023</mn><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>23</mn></mrow></mrow></msup></mrow></mrow></mfrac></math> <math> <mo>×</mo> </math> 7.884 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>28</mn></mrow></mrow></msup></math> = 30.76 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msup></math> kgHence, the Uranium needed per year is 30.76<math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msup></math> kg

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