HW6_solution-1

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Homework 6, Solution 8.2. Revisit Example 8-2. (a) If the cost of raw uranium is $65/kg, what percent of the cost of $40 million worth of fuel rods does this amount constitute? (b) If the cost increases to $260/kg, and all other cost components of the refueling remain the same, what is the new cost of the refueling each year? Solution: From the example, the mass of the fuel rods is 155,000 kg, of which one-third is replaced each year, or 51,667 kg. Part (a): Since the cost is $65/kg, the cost of replacing 51,667 kg is $3.358 million, or 8.4% of $40 million. Part (b): With the cost increased to $260/kg, the cost of replacing 51,667 kg is now $13.43 million, or 26.8% of $40 million. Comment: The calculation shows that the overall cost of electricity from fission of uranium is probably not very sensitive to potential future increases in the cost of uranium supplies. This is indicated because even at the highest predicted price level, the uranium constitutes only 27% of the cost of the fuel rods, which are in turn only part of the overall levelized cost per kWh of electricity. 8.3. Repeat the calculation of levelized cost per kWh for electricity in Example 8-2 applied to a new nuclear plant with a total capital cost of $2200/kW, an output of 1000 MW e , an investment lifetime of 20 years, and an MARR of 5%. This is the cost of one of the most recently completed nuclear plants in the United States, at Seabrook, NH, according to one estimate. Include fuel, O&M, and capital cost in your calculation. Solution: For this plant, it is assumed that the capacity factor is 90%, so that the annual output is:   year GWh yr h MW / 884 , 7 % 90 8760 1000          The capital cost of the plant is $2.2 billion. Therefore, the annual capital cost can be calculated, e.g., using the spreadsheet function –PMT: AnnCapCost = –PMT(5%, 20, $2.2B) = $176.5M From the example in the chapter, operation and maintenance cost is $118M per year, and fuel cost is $0.00432/kWh. Based on the number of kWh produced per year, the total fuel cost is $34.1M. Levelized cost per kWh is therefore kWh kWh M M M / 0417 . 0 $ 10 884 . 7 1 . 34 $ 118 $ 5 . 176 $ 9     8.4. Carry out the rudimentary design of a small modular nuclear reactor that delivers 100 MW e . The reactor is based on a Rankine cycle where the steam is expanded in a turbine from 12 MPa and 330°C to condensed water at 5 kPa. Assume that the generator is 98% efficient, the actual cycle achieves 90% of the efficiency of the ideal cycle, and 90% of

the thermal energy released in the nuclear reactions is transferred to the working fluid. (a) If the reactor uses uranium as a fuel which is enriched to 3% U-235, what mass of fuel is consumed each year? (b) What is the flow rate of steam in kg/s required when the reactor is running at full power? Solution: Part (a): To calculate fuel consumption, it is first necessary to make all needed computations for the ideal thermodynamic cycle. Pump work from state 1 to 2 based on change in pressure and specific volume is 12.6 kJ/kg. Based on the quality of the steam at the turbine exit, the enthalpy is 1740.4 kJ/kg. The following table of values can then be compiled from thermodynamic tables and subsequent calculations: State h s (kJ/kg) (kJ/kgK) 1 137.8 0.4762 2 150.4 * 3 2826.6 5.713 4 1740.4 5.713 From the table, the heat added is 2676.2 kJ/kg, and the turbine output is 1086.2 kJ/kg, so the ideal efficiency is: % 1 . 40 2 . 2676 6 . 12 2 . 1086      in p t th Q w w  The actual thermodynamic efficiency is 90% of this amount, or 36.1%. Taking into account losses in the transfer of thermal energy from the reaction to the fluid, and in the generator, the overall efficiency is: η overall = (0.9)(0.361)(0.98) = 31.8% If the average output is 100 MW, then based on overall efficiency, the average input is 314 MW. Thereafter, the amount of 3% enriched uranium fuel is calculated as   U kg 4729 % 3 1 8 . 69 1 0036 . 0 8760 314                            GJ kg MWh GJ yr h MW Since the uranium fuel is contained in UO 2 , the total mass of fuel is 5365 kg/yr. Part (b): To generate 100 MW e , taking into account generator losses, the turbine must product 102.04 MW. Since the net turbine work is 1.0737 MJ/kg, the net steam flows is

sec / 95 0737 . 1 0204 . 1 kg  8.6. A reactor has a rated capacity of 1 GW e and produces 7.9 billion kWh/year. It is fueled using fuel rods that contain 2.5% U-235 by mass. The electric generator is 98% efficient, the thermal cycle of the plant is 33% efficient and 90% of the thermal energy released in fission is transferred to the working fluid. (a) What is the capacity factor of the plant? (b) What is the rate of U-235 consumption in g/s in order to deliver the average output? (c) What is the total mass of fuel rods consumed each year? (d) Based on the amount of energy released in fission reactions, what is the reduction in mass of fuel each year? Solution: Part (a): The capacity factor is the actual output divided by hypothetical output if the plant had operated at maximum capacity for the entire year:   % 2 . 90 / 10 76 . 8 / 10 9 . 7 / 8760 10 1 / 10 9 . 7 9 9 6 9       yr kWh yr kWh yr h kW yr kWh Part (b): Fuel consumption is based on an energy content of 69.8 TJ/kg U-235. The average power output of the plant is 902 MW e . The average energy input from the fuel incorporating all losses is calculated as follows:   s J s MJ MW MW / 10 098 . 3 / 3098 3098 9 . 0 1 33 . 0 1 98 . 0 1 902 9                       Fuel consumption is then a function of rate of energy consumption and energy content per mass of fuel: s g kg J s J / 044 . 0 / 10 98 . 6 / 10 098 . 3 10 9    Part (c): Mass of fuel rod consumption per year incorporates the enrichment level of 2.5%:     yr kg kgU kgFuel yr s s g / 995 , 55 235 025 . 0 1 / 10 1536 . 3 / 044 . 0 7           Assuming the fuel is in the form of UO 2 (uranium oxide), the total mass is 63,524 kg/yr. Part (d): The energy input per year from fission is calculated as follows:   yr J yr s s J / 10 77 . 9 / 10 1536 . 3 / 10 098 . 3 16 7 9     Using Einstein’s formula gives the reduction in mass:

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kg J c E m 09 . 1 10 9 10 77 . 9 16 16 2      Thus, the reaction leads to the conversion of 1.09 kg of mass to energy each year. 8.8. Suppose that the Yucca Mountain facility discussed in the chapter ends up costing $60 billion, and stores 70,000 tonnes of nuclear waste. Ignoring transportation costs for waste, and the effect of the time value of money, what is the surcharge that is required on each kWh of electricity to exactly pay for the facility, based on the amount of electricity production that results in this volume of waste? State your assumptions regarding the amount of waste generated per unit of electricity produced. Solution: For the purposes of this solution, we assume that a typical plant produces 9255 GWh per year, which results in 58,600 kg of waste per year. Accordingly, the rate of waste production is: kWh g yr kWh yr g / 00633 . 0 / 10 255 . 9 / 10 86 . 5 9 7    The total amount of electricity production that can be supported by Yucca Mountain is therefore: kWh kWh g g 13 10 10 106 . 1 / 00633 . 0 10 7    Dividing total cost by electricity production gives the cost per kWh: kWh kWh B / 00543 . 0 $ 10 106 . 1 60 $ 13  