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Chapter 15: Problem 29

If a nuclear plant is built for \(\$ 10\) billion and operates for 50 years under an operating cost of \(\$ 100\) million per year, what is the cost to produce electricity, in \(\$ / \mathrm{kWh}\) assuming that the plant delivers power at a steady rate of \(1 \mathrm{GW}\) for the whole time?

Short Answer

Expert verified
Answer: The cost to produce electricity by the nuclear plant is approximately $0.0343 per kWh.

Step by step solution

01

Calculate the Total Cost of the Nuclear Plant

To find the total cost of the nuclear plant, we need to consider both the initial investment cost and the operating costs over 50 years. The initial investment cost is given as \(\$10\) billion. The annual operating cost is given as \(\$100\) million per year. So, the total operating cost for 50 years is \(\$100\) million/year * 50 years = \(\$5\) billion. Now, we can add the initial investment and the total operating cost to get the total cost: Total Cost = \(\$10\) billion + \(\$5\) billion = \(\$15\) billion.
02

Convert the Power Delivery Rate to kWh

The power delivery rate of the nuclear plant is given as \(1\) GW. Since we need to find the cost per kWh, we will convert this power rate into kWh. We know that \(1\) GW = \(1,000,000\) kW. Thus, the power delivery rate is \(1,000,000\) kW.
03

Calculate the Total Operating Hours

We are given that the nuclear plant operates for 50 years. First, we will calculate the total operating hours during this period: Total Hours = 50 years * 365 days/year * 24 hours/day = 438,000 hours.
04

Calculate the Total kWh Produced

Now that we have the power delivery rate in kWh and the total operating hours, we can calculate the total kWh produced by the nuclear plant during its 50-year operation: Total kWh = Power Delivery Rate * Total Operating Hours = 1,000,000 kW * 438,000 hours = 438,000,000,000 kWh.
05

Calculate the Cost per kWh

Finally, to find the cost per kWh, we need to divide the total cost of the nuclear plant by the total kWh produced: Cost per kWh = Total Cost / Total kWh = \(\frac{\$15,000,000,000}{438,000,000,000 \ \mathrm{kWh}} = 0.0343 \ \$/\mathrm{kWh}\). So, the cost to produce electricity by this nuclear plant is approximately \(0.0343\) per kWh.

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Most popular questions from this chapter

Explain in some detail what happens if control rods are too effective at absorbing neutrons so that each fission event produces too few unabsorbed neutrons.

Which of the following is true about the fragments from \(\mathrm{a}^{235} \mathrm{U}\) fission event? a) any number of fragments ( 2 through 235 ) can be produced b) a small number of fragments will emerge ( 2 to 5 ) c) two nearly identical fragments will emerge d) two fragments of distinctly different size will emerge e) the fission is an alpha decay: a small piece having \(A=4\) is emitted

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?

In close analog to the half-lives of \({ }^{235} \mathrm{U}\) and \({ }^{238} \mathrm{U}\), let's say two 80 elements have half lives of \(4.5\) billion years and 750 million years. If we start out having the same number of each (1:1 ratio), what will the ratio be after \(4.5\) billion years? Express as \(x: 1\), where \(x\) is the larger of the two.

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?
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