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It is proposed to use nuclear fusion reaction : $\; _{1}H^{2}+_{1}H^{2}=\;_{2}He^{4}\;$ in a nuclear reactor of 200MW rating . If energy from above reaction is used with a $\;25 \%\;$ efficiency in reactor how many gr ams of deuterium will be needed per day . (The masses of $\;_{1}H^{2}\;$ and $\;_{2}He^{4}\;$ are 2.0141 & 4.0026 amu respectively

$(a)\;121 gm/day\qquad(b)\;112 gm/day\qquad(c)\;128 gm/day\qquad(d)\;100 gm/day$

1 Answer

Answer : (a) $\;121 gm/day$
Explanation :
Let us first calculate Q value of nuclear function
$Q=\bigtriangleup mC^2=\bigtriangleup m\;(931) MeV$
$Q=(2\times2.0141-4.0026)\times931MeV$
$=23.834MeV=23.834\times10^{6}eV$
Now efficiency of reactor is $\;25 \%$
$=\large\frac{25}{100}\times23.834\times10^{6}\times1.6\times10^{-19} J$
$=9.534 \times10^{-13} J$
Now $\;9.534\times10^{-13} J\;$ energy is released by fusion of 2 deuterium
$\large\frac{9.534\times10^{-13}}{2} J/deuterium\;$ is released Requirement is $\;200MW=200 \times10^{6} J/s\times86400\;$ for 1 day
no . of deuterium nuclei required = $\;\large\frac{200\times10^{6}\times86400}{\large\frac{9.534}{2}\times10^{-13}}$
$=3.624\times10^{25}$
No.of deucterium nuclei
$=\large\frac{g}{A}\times6\times10^{23}$
$3.624\times10^{25}=\large\frac{g}{2}\times6\times10^{23}$
$g=2\times\large\frac{3.624\times10^{25}}{6\times10^{23}}=120.83 gm/day\;.$
answered Feb 24, 2014 by yamini.v
 

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