A nuclear power plant, whose capital cost has been fully repaid, operates with a total of 155,000 kg of uranium (or 175,800 kg of UO2) in its reactor core, which has a U-235 concentration of 2.8%. Each year 1/3 of the reactor fuel is replaced at a cost of $40 million, which has built into it $0.001

Question

A nuclear power plant, whose capital cost has been fully repaid, operates with a total of 155,000 kg of uranium (or 175,800 kg of UO[latex]_2[/latex]) in its reactor core, which has a U-235 concentration of 2.8%. Each year 1/3 of the reactor fuel is replaced at a cost of $40 million, which has built into it $0.001/kWh to be paid to a fund for the management of nuclear waste. The facility also incurs $118 million/year in nonfuel operating and maintenance cost.

The working fluid is compressed to 10 MPa, and then heated to 325°C, after which it is expanded in the turbine to 10 kPa, condensed, and returned to the pump for return to the reactor core. The generator is 98% efficient. Assume that all energy released in nuclear fission is transferred to the working fluid, and that, due to various losses, the actual Rankine cycle used achieves 90% of its ideal efficiency. Ignore the contribution of fission of Pu-239 that results from transformation of some of the U-238 in the fuel. Calculate the cost per kWh of production for this plant.

Accepted Answer

We solve this problem by first calculating the thermal efficiency of the cycle and the rate of fuel consumption, and then the annual electrical output and the components of total cost.

From the steam tables, at 10 kPa we have h[latex]_f[/latex] = 191.8 kJ/kg, h[latex]_g[/latex] = 2584.7 kJ/kg, s[latex]_f[/latex] = 0.6493 kJ/kg·K, and s[latex]_g[/latex] = 8.1502 kJ/kg·K. Using the standard technique for analyzing an ideal Rankine cycle from Chap. 6 gives ω[latex]_{pump}[/latex] = 10.1 kJ/kg and the following table of temperature, enthalpy, and entropy values at states 1 to 4:

State	Temp. (C)	Enthalpy (h) (kJ/kg)	Entropy (s) (kJ/kg · K)
1	45.8	191.8	0.6493
2	*	201.9	*
3	320	2809.1	5.7568
4	*	1821.1	5.7568

∗Value is not used in the calculation, and therefore not given

Based on the enthalpy values h[latex]_1[/latex] – h[latex]_4[/latex], we obtain q[latex]_{in}[/latex] = 2607.2 kJ/kg, and ω[latex]_t[/latex] = 988.0 kJ/kg. Accordingly, the actual efficiency of the cycle is the ideal efficiency multiplied by the cycle and generator losses:

[latex]\eta=\frac{988.0-10.1}{2607.2}(0.9)(0.98)=33.1 \%[/latex]

Next, regarding the fuel, since one third of the fuel is replaced each year, the annual fuel consumption is one-third of the total, or 51,700 kg. Taking into account the proportion of the fuel that is U-235, the consumption rate m is

[latex]m=\frac{\left(51.7 imes 10^{6} \mathrm{~g} ight)(0.028)}{3.154 imes 10^{6} \mathrm{~s}}=0.0459 \mathrm{~g} / \mathrm{s}[/latex]

Given an energy release of 69.8 TJ/1 kg or 69.8 GJ/g of fissioned U-235, the consumption of fuel is transformed into electrical output as follows:

(input) × (effi ciency) × (h/year) = output

[latex](0.0459 \mathrm{~g} / \mathrm{s})\left(69.8 \frac{\mathrm{GJ}}{\mathrm{g}} ight)(0.331)\left(1 \frac{\mathrm{GW}}{\mathrm{GJ} / \mathrm{s}} ight)\left(8760 \frac{\mathrm{h}}{ ext { year }} ight)=9279 \mathrm{GWh} / ext { year }[/latex]

The average output of the plant is 1.059 GW[latex]_e[/latex]. The peak output of the plant is not known exactly from the data given, but is on the order of 1.2 GW[latex]_e[/latex], to take into account downtime and the fluctuations in demand for output.
Total cost per kWh is calculated from the fuel and O&M cost components. Fuel cost is ($40 million)/(9.255 × 10[latex]^9[/latex] kWh) = $0.00431/kWh. O&M cost is ($118 million)/(9.255 × 10[latex]^9[/latex] kWh) = $0.0127/kWh. Thus the total cost is $0.0170/kWh

Discussion The result of the calculation shows the cost-competitiveness of a fully amortized nuclear power plant. The cost of fuel and O&M are at or very close to the U.S. average values of $0.0045/kWh and $0.0127/kWh, published by the U.S.-based Nuclear Energy Institute. At $65/kg, the cost of the raw uranium constitutes only a small fraction of the total cost of the fuel resupply, so even a substantial increase in this price would not appreciably change the overall cost of electricity production. Actual plants might vary somewhat from the value given, for example if the maximum pressure is different, or if one takes into account the contribution from fissioning of Pu-239, especially toward the end of a fuel rod’s time in the reactor core when its concentration is highest. Also, not all of the losses of efficiency have been detailed in this simplified problem, so the electrical output per unit of fuel consumed is slightly overstated. However, these modifications to the calculation would not appreciably change its outcome, namely, that the cost of electricity production is approximately $0.02/kWh.
A more substantial impact on electricity cost would come from the recovery of capital investment (the calculation appears as an exercise at the end of the chapter). Up until the rise in fossil fuel prices that started around 2005, it was difficult to justify the capital cost of nuclear power, despite the large amount of energy that could be produced per dollar spent on fuel. In a number of countries, especially those with large domestic coal resources such as the United States or China, investors had been reluctant in recent years to invest in new plants for this reason. Some countries with few domestic energy resources, such as France and Japan, have continued to build nuclear plants, judging that the energy security benefit for a country completely dependent on imported fossil fuel outweighed any additional cost for nuclear compared to fossil.

Question 8.2: A nuclear power plant, whose capital cost has been fully rep...

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