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1kg fuel is consumed in 20 days in U-238 nuclear reactor. If the useful energy per fission is 190 MeV,then the output powet of the reactor will be? a- 15MW b-30MW c-45MW d-60MW?
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1kg fuel is consumed in 20 days in U-238 nuclear reactor. If the usefu...
Given Information:
- 1 kg of fuel is consumed in 20 days in U-238 nuclear reactor.
- The useful energy per fission is 190 MeV.
To find:
The output power of the reactor.
Formula:
The output power of a nuclear reactor can be calculated using the following formula:
Output Power = (Energy released per fission) × (Number of fissions per unit mass) × (Mass of fuel consumed per unit time)
Solution:
Step 1: Calculate the number of fissions per unit mass
To calculate the number of fissions per unit mass, we need to know the molar mass of U-238 and Avogadro's number.
Molar mass of U-238 = 238 g/mol
Avogadro's number = 6.022 × 10^23 atoms/mol
The number of U-238 atoms in 1 kg of fuel can be calculated as follows:
Number of atoms = (Mass of fuel in grams) / (Molar mass of U-238)
Number of atoms = (1000 g) / (238 g/mol)
Number of atoms = 4201.68 × 10^23 atoms
Since each U-238 atom undergoes fission, the number of fissions per unit mass is equal to the number of atoms in the fuel.
Number of fissions per unit mass = 4201.68 × 10^23 fissions/kg
Step 2: Calculate the mass of fuel consumed per unit time
Mass of fuel consumed per unit time = (Mass of fuel consumed) / (Time)
Mass of fuel consumed per unit time = (1 kg) / (20 days)
Mass of fuel consumed per unit time = 0.05 kg/day
Step 3: Calculate the output power
Output Power = (Energy released per fission) × (Number of fissions per unit mass) × (Mass of fuel consumed per unit time)
Energy released per fission = 190 MeV = 190 × 10^6 eV
1 eV = 1.602 × 10^-19 Joules
Energy released per fission = 190 × 10^6 × 1.602 × 10^-19 J
Output Power = (190 × 10^6 × 1.602 × 10^-19 J) × (4201.68 × 10^23 fissions/kg) × (0.05 kg/day)
Output Power = (190 × 1.602 × 4201.68 × 0.05) × (10^6 × 10^-19 × 10^23) J/day
Output Power = 0.63 × 10^10 J/day
Since the unit of power is Watts, we need to convert the output power from Joules to Watts.
1 Watt = 1 Joule/second
1 day = 24 hours = 24 × 60 × 60 seconds
Output Power = (0.63 × 10^10 J/day) / (24 × 60 × 60 s/day)
Output Power = 730.9028 Watts
Step 4: Convert the output power to megawatts
1 Megawatt = 10^6 Watts
Output Power
- 1 kg of fuel is consumed in 20 days in U-238 nuclear reactor.
- The useful energy per fission is 190 MeV.
To find:
The output power of the reactor.
Formula:
The output power of a nuclear reactor can be calculated using the following formula:
Output Power = (Energy released per fission) × (Number of fissions per unit mass) × (Mass of fuel consumed per unit time)
Solution:
Step 1: Calculate the number of fissions per unit mass
To calculate the number of fissions per unit mass, we need to know the molar mass of U-238 and Avogadro's number.
Molar mass of U-238 = 238 g/mol
Avogadro's number = 6.022 × 10^23 atoms/mol
The number of U-238 atoms in 1 kg of fuel can be calculated as follows:
Number of atoms = (Mass of fuel in grams) / (Molar mass of U-238)
Number of atoms = (1000 g) / (238 g/mol)
Number of atoms = 4201.68 × 10^23 atoms
Since each U-238 atom undergoes fission, the number of fissions per unit mass is equal to the number of atoms in the fuel.
Number of fissions per unit mass = 4201.68 × 10^23 fissions/kg
Step 2: Calculate the mass of fuel consumed per unit time
Mass of fuel consumed per unit time = (Mass of fuel consumed) / (Time)
Mass of fuel consumed per unit time = (1 kg) / (20 days)
Mass of fuel consumed per unit time = 0.05 kg/day
Step 3: Calculate the output power
Output Power = (Energy released per fission) × (Number of fissions per unit mass) × (Mass of fuel consumed per unit time)
Energy released per fission = 190 MeV = 190 × 10^6 eV
1 eV = 1.602 × 10^-19 Joules
Energy released per fission = 190 × 10^6 × 1.602 × 10^-19 J
Output Power = (190 × 10^6 × 1.602 × 10^-19 J) × (4201.68 × 10^23 fissions/kg) × (0.05 kg/day)
Output Power = (190 × 1.602 × 4201.68 × 0.05) × (10^6 × 10^-19 × 10^23) J/day
Output Power = 0.63 × 10^10 J/day
Since the unit of power is Watts, we need to convert the output power from Joules to Watts.
1 Watt = 1 Joule/second
1 day = 24 hours = 24 × 60 × 60 seconds
Output Power = (0.63 × 10^10 J/day) / (24 × 60 × 60 s/day)
Output Power = 730.9028 Watts
Step 4: Convert the output power to megawatts
1 Megawatt = 10^6 Watts
Output Power
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1kg fuel is consumed in 20 days in U-238 nuclear reactor. If the usefu...
C- 45MW
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1kg fuel is consumed in 20 days in U-238 nuclear reactor. If the useful energy per fission is 190 MeV,then the output powet of the reactor will be? a- 15MW b-30MW c-45MW d-60MW?
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1kg fuel is consumed in 20 days in U-238 nuclear reactor. If the useful energy per fission is 190 MeV,then the output powet of the reactor will be? a- 15MW b-30MW c-45MW d-60MW? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 1kg fuel is consumed in 20 days in U-238 nuclear reactor. If the useful energy per fission is 190 MeV,then the output powet of the reactor will be? a- 15MW b-30MW c-45MW d-60MW? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1kg fuel is consumed in 20 days in U-238 nuclear reactor. If the useful energy per fission is 190 MeV,then the output powet of the reactor will be? a- 15MW b-30MW c-45MW d-60MW?.
1kg fuel is consumed in 20 days in U-238 nuclear reactor. If the useful energy per fission is 190 MeV,then the output powet of the reactor will be? a- 15MW b-30MW c-45MW d-60MW? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 1kg fuel is consumed in 20 days in U-238 nuclear reactor. If the useful energy per fission is 190 MeV,then the output powet of the reactor will be? a- 15MW b-30MW c-45MW d-60MW? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1kg fuel is consumed in 20 days in U-238 nuclear reactor. If the useful energy per fission is 190 MeV,then the output powet of the reactor will be? a- 15MW b-30MW c-45MW d-60MW?.
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