Question

What is the power output of $_{92}{U^{235}}$ reactor if it takes $30$ days to use up $2kg$ of fuel and if each fission gives $185MeV$ of usable energy? Avogadro's number $= 6.02 \times {10^{26}}Kmol$ .
A) $56.3MW$
B) $60.3MW$
C) $58.3MW$
D) $54.3MW$

Answer 1

In the question, we have to find the power output of the reactor. To find power, we need to find the energy released by the reactor in $30$ days. For this we can use the unitary method. The energy for each fission which means energy from each atom is already given in the question. We need to find how much the mass of each atom is and use it to find the answer.

Formulae used:
$1amu = 1.66 \times {10^{ - 27}}kg$
$1MeV = 1.6 \times {10^{ - 13}}J$
$P = \dfrac{E}{t}$
Here $P$ is the power, $E$ is the energy in joules and $t$ is the time in seconds.

Complete step by step solution:
In the question it is given that each nuclear fission gives $185MeV$ amount of usable energy.
This means,
$\Rightarrow$ $235amu$ of uranium gives $185MeV$ energy.
$\Rightarrow$ $1amu$ of uranium gives $\dfrac{{185}}{{235}}MeV$ energy.
Let this be equation 1.
We know that $1amu = 1.66 \times {10^{ - 27}}kg$
Also $1MeV = 1.6 \times {10^{ - 13}}J$
Using these values in equation 1, we get
$\Rightarrow$ $1.66 \times {10^{ - 27}}kg$ of uranium gives $\dfrac{{185}}{{235}} \times 1.6 \times {10^{ - 13}}J$ energy.
$\Rightarrow$ $1.66 \times {10^{ - 27}}kg$ of uranium gives $1.259 \times {10^{ - 13}}J$ energy.
$\Rightarrow$ $1kg$ of uranium gives $\dfrac{{1.259 \times {{10}^{ - 13}}}}{{1.66 \times {{10}^{ - 27}}}}J$ energy.
$\Rightarrow$ $2kg$ of uranium gives $\dfrac{{1.259 \times {{10}^{ - 13}}}}{{1.66 \times {{10}^{ - 27}}}} \times 2J$ energy.
$\Rightarrow$ $2kg$ of uranium gives $\dfrac{{1.259 \times {{10}^{ - 13}}}}{{1.66 \times {{10}^{ - 27}}}} \times 2J$ energy.

$\Rightarrow$ $2kg$ of uranium gives $1.51 \times {10^{14}}J$ energy.
So the usable energy produced by the reactor after consuming $2kg$ of uranium will be $1.51 \times {10^{14}}J$ .
We know that,
$\Rightarrow P = \dfrac{E}{t}$
Let this be equation 2.
Here $P$ is the power produced by the reactor, $E$ is the useful energy released by the reactor in joules and $t$ is the time (in seconds) for which we have to calculate the power.
In the question time given is 30 days. The time in seconds will be,
$\Rightarrow t = 30 \times 24 \times 60 \times 60\sec$
$\Rightarrow t = 2592000\sec$

Substituting the values of $E$ and $t$ in equation 2, we get the power produced by the reactor to be,

$\Rightarrow P = \dfrac{{1.51 \times {{10}^{14}}J}}{{2592000\sec }} = 5,82,56,172.8W \approx 58.3MW$

So the answer will be option (C).

Note: While solving the above question, we have to be very useful with the units. The mass of individual atoms is always taken in atomic mass units. Also, while calculating power we have to take care of the unit of time because power is always calculated with respect to time in seconds. So we have to convert the given time in seconds.