Question

A town has a population of 1 million. The average electric power needed per person is 300 W. A reactor is to be designed to supply power to this town. The efficiency with which thermal power is converted into electric power is aimed at 25%. (a) Assuming 200 MeV to thermal energy to come from each fission event on an average, find the number of events that should take place every day. (b) Assuming the fission to take place largely through 235U, at what rate will the amount of 235U decrease? Express your answer in kg per day. (c) Assuming that uranium enriched to 3% in 235U will be used, how much uranium is needed per month (30 days)?

Answer 1

Answer:

fyi and your family are you mad not i am giving questions about how much more than a project on

Explanation:

5ttyyttrtrftyudjfd8userid and your family are you mad not i am giving questions about how much more than a project on French and remote areas of the day god bless

Answer 2

a. The number of events that should take place every day $N=3.24 \times 10^{24}$

b. The rate at which the amount of 235 U decrease $=1.2643 \mathrm{kg}$

Explanation:

(a) The town’s total population = 1 million =  106

On an average, the electric power required per person = 300 W

In one day, the total power used = $300 \times 10^{6} \times 60 \times 60 \times 241=86400 \times 300 \times 10^{6} \mathrm{J}$

In one fission, the generated energy $=200 \times 1.6 \times 10^{6} \times 10^{-19} \mathrm{J}=3.2 \times 10^{-11} \mathrm{J}$

25% is the efficiency with which the thermal power is transformed into electric power.

So, the electrical energy is shown as $E=8 \times 10^{-12} J$

Let us consider that the fission is denoted as N.

Hence, N fissions’ total energy $=N \times 8 \times 10^{-12}$

According to the provided data,

$8 \times 10^{-12} \times N=86400 \times 300 \times 10^{6} J$

$N=3.24 \times 10^{24}$

(b) Per day, the required number of moles $n=\mathrm{N} 6.023 \times 1023$

⇒n$=5.38 \mathrm{mol}$

So, the required amount of uranium per day =$5.38 \times 235=1.2643 \mathrm{kg}$

(c) The required amount of uranium in a month =$1.264 \times 30 \mathrm{kg}$

In 235U, let us consider that x kg of uranium enhanced to 3% be used.

⇒ $x \times 3100$ ⇒ $x=1264 \mathrm{kg}$