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Chapter 15: Problem 30

Since each nuclear plant delivers \(\sim 1 \mathrm{GW}\) of electrical power, at \(\sim 40 \%\) thermodynamic efficiency this means a thermal generation rate of \(2.5\) GW. How many nuclear plants would we need to supply all 18 TW of our current energy demand? Since a typical lifetime is 50 years before decommissioning, how many days, on average would it be between new plants coming online (while old ones are retired) in a steady state?

Short Answer

Expert verified
Answer: 7200 nuclear power plants are required to supply the current energy demand of 18 TW. In a steady state, there would need to be a new nuclear power plant commissioned every 2.53 days while old ones are retired.

Step by step solution

01

Calculate the number of nuclear plants required to meet the energy demand

We are given the electrical power output of each nuclear plant as 1 GW. The thermodynamic efficiency of each plant is 40%, which means that they generate thermal power at a rate of 2.5 GW. We want to find out how many nuclear plants are needed to supply the current energy demand of 18 TW. To do this, we'll divide the total energy demand by the thermal power generation rate of a single nuclear plant: Number of nuclear plants = Total energy demand / Thermal power generation rate per plant Number of nuclear plants = \((18 \times 10^{12} \,\mathrm{W}) \div (2.5 \times 10^{9} \,\mathrm{W})\)
02

Calculate the number of nuclear plants needed

Now we can calculate the Number of nuclear plants: Number of nuclear plants = \((18 \times 10^{12} \,\mathrm{W}) \div (2.5 \times 10^{9} \,\mathrm{W}) = 7200\) So, 7200 nuclear power plants are required to supply the current energy demand of 18 TW.
03

Determine the rate of new plants coming online

Given that a typical nuclear plant has a lifetime of 50 years, we can calculate the rate at which new plants need to come online to maintain a steady state. We will divide the total number of plants by their lifetime (in days) to find out how many plants need to be commissioned per day: Rate of new plants coming online = Total number of nuclear plants / Plant lifetime (in days) Plant lifetime (in days) = 50 years \(\times\) 365 days/year = 18250 days Rate of new plants coming online = 7200 nuclear plants / 18250 days
04

Calculate the average number of days by which new plants need to come online

Now, we can find the average number of days between new plants coming online: Average number of days between new plants = Total plant lifetime (in days) / Number of nuclear plants = 18250 days / 7200 Average number of days between new plants ≈ 2.53 days In a steady state, there would need to be a new nuclear power plant commissioned every 2.53 days while old ones are retired.

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Most popular questions from this chapter

In a year, an average American uses about \(3 \times 10^{11} \mathrm{~J}\) of energy. How much mass does this translate to via \(E=m c^{2} ?\) Rock has a density approximately 3 times that of water, translating to about \(3 \mathrm{mg}\) per cubic millimeter. So roughly how big would a chunk of rock material be to provide a year's worth of energy if converted to pure energy? Is it more like dust, a grain of sand, a pebble, a rock, a boulder, a hill, a mountain?

Operating approximately 450 nuclear plants over about 60 years at a total thermal level of \(1 \mathrm{TW}\), we have had two major radioactive releases into the environment. If we went completely down the nuclear road and get all \(18 \mathrm{TW}^{89}\) this way, what rate of accidents might we expect, if the rate just scales with usage levels?

Control rods in nuclear reactors tend to contain \({ }^{10} \mathrm{~B}\), which has a high neutron absorption cross section. \(^{81}\) What happens to this nucleus when it absorbs a neutron, and is the result stable? If not, track the decay chain until it lands on a stable nucleus.

A large boulder whose mass is \(1,000 \mathrm{~kg}\) having a specific heat capacity of \(1,000 \mathrm{~J} / \mathrm{kg} /{ }^{\circ} \mathrm{C}\) is heated from \(0^{\circ} \mathrm{C}\) to a glowing \(1,800^{\circ} \mathrm{C}\). How much more massive is it, assuming no atoms have been added or subtracted?

A particular fission of \({ }^{235} \mathrm{U}+\mathrm{n}(\) total \(A=236)\) breaks up. One fragment has \(Z=54\) and \(N=86\), making it \({ }^{140} \mathrm{Xe} .\) If no extra neutrons are produced in this event, what must the other fragment be, so all numbers add up? Refer to a periodic table (e.g., Fig. B.1; p. 375 ) to learn which element has the corresponding \(Z\) value, and express the result in the notation \({ }^{\mathrm{A}} \mathrm{X}\).
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