Nuclear Power Plants - Electricity Production

In summary, based off of the equation given above, a single atom of uranium-235 releases 2.5977713481*10^-11 Joules of energy when fissioned. This amount of energy can be converted into other units such as watts, kilowatts, or petawatts. In order to create 2.49 Peta Watts of electricity, a mass of uranium 235 would need to be fissioned.
  • #1
eminem4002
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Homework Statement
An average efficiency of a nuclear power plant is 35%. How much mass of uranium-235 is required to produce 2.49 PWh of electric energy?
Relevant Equations
Uranium-235 + neutron -> Krypton-92 + Barium-141 + 3 neutrons.
I've calculated that reaction-energy in fission of a single atom of Uranium-235 is 2.5977713481*10^-11 Joules (based off of the equation given above.) I'm assuming that PWh has to be converted into simpler units so, 2.49PWh = 2.49*10^15Wh.. afterwards I could make it into watts but I was never given a specific time period in which the uranium has to produce 2.49 PWh of electricity. Am I overthinking this, maybe I completely missed the point of the task?
 
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  • #2
It doesn't matter. PWh is a unit of energy. (As is Wh)
2.49 PWh could be (in abstract) 2.49 PW for 1 hour, or 2.49 GW for 10^6 hours or 28.4 GW for 10 years, etc. (E&OE)
 
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  • #3
eminem4002 said:
Am I overthinking this, maybe I completely missed the point of the task?
Please, see:
https://en.wikipedia.org/wiki/Kilowatt_hour

The electric bills you have been paying have been telling you how much energy (Kw-h) you have consumed within a month.
They don't care if you used all in one single day of that month.

I don't have a clue about how much energy you can obtain from the fission of a single atom of Uranium-235. :frown:
 
  • #4
Hey, thanks for the suggestions guys. I've figured out that the energy released per uranium atom may variate quite a bit based on the sources where you find the atomic weight from. I used https://periodictable.com/Isotopes/056.141/index.html to get my numbers and got 2.5977713481*10^-11 Joules per uranium-235 atom and looking at Wikipedia page of uranium-235 (https://en.wikipedia.org/wiki/Uranium-235#cite_note-kayelaby-2 ) the number happens to be 3.24×10^−11 Joules. However, I'm still stuck on the main task at hand...
 
  • #5
You did the hard part! You now know how many Joules you get from an atom of U235. All you have to do is work out how many Joules is the same as 2.49 PWh and see how many atoms you need.

Eg. I 1 atom gives 3 x 10^-11 J and you need 6 x 10^10 J, then you need ##\frac {6 \times 10^{10}}{ 3 \times 10^{-11}} \, ## atoms (at 100% efficiency.)

If you can't convert PWh to Joules, you need to remember a Watt is 1 Joule per second, or a Jouls is 1 W for 1 sec.
A Wh is 1 W for 1 hour = 1W for ? sec = ? J
etc.
 
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  • #6
I first converted 2.49 PWh to Joules and got 8.964*10^18 Joules. Divided this value by 0.35 as that's the efficiency of the power plant and got 2.561142857143*10^19 Joules. As mentioned earlier I found out that single a fission makes 2.5977713481*10^-11 Joules. Divide the required energy by the amount of energy created by a single fission ( 2.561142857143*10^19 Joules / 2.5977713481*10^-11 Joules) and got 9.859000327401*10^29 fissions. From there on I found out the mass of uranium in kilograms per fission. ( 235.043929918u * 1.66*10^-27kg = 3.901729236639*10^-25 kg per fission. The 1.66*10^-27 kg is to convert the mass of the nucleus into kilograms. At the end, I multiplied this value by how many fissions there are in total. ( 3.901729236639*10^-25 kg * 9.859000327401*10^29 fissions = 384671.4982145 kg... 384671.4982145kg is the required mass of uranium to achieve 2.49 PWh at one point. I really feel like I did something wrong, can someone show me where I went wrong?
 
  • #7
The calculations all look correct to me. All are well reasoned and explained. Though you've carried a lot of insignificant digits into the result.
 
  • #8
jbriggs444 said:
The calculations all look correct to me. All are well reasoned and explained. Though you've carried a lot of insignificant digits into the result.
I was told to do so when calculating things related to nuclear physics. However does it seem logical that it would take so many kilograms of uranium 235 to create 2.49 Peta Watthours? I know that peta is a huge number but would it really take so much mass to create this much electricity?
 
  • #9
eminem4002 said:
I was told to do so when calculating things related to nuclear physics. However does it seem logical that it would take so many kilograms of uranium 235 to create 2.49 Peta Watthours? I know that peta is a huge number but would it really take so much mass to create this much electricity?
A quick trip to Google with a search term of "how much uranium per petawatt" turns up a hit that does not answer the question directly but does tell you that it is a large mass.
Google said:
1 kilogram of enriched uranium will produce about 325 megawatt hours of electricity in a modern nuclear power plant. This is a small fraction of the actual energy in uranium because current nuclear reactors trade safety for efficiency.
 
  • #10
jbriggs444 said:
A quick trip to Google with a search term of "how much uranium per petawatt" turns up a hit that does not answer the question directly but does tell you that it is a large mass.
Okay so I also searched up and you're right! Thank you for the assistance!
 
  • #11
jbriggs444 said:
A quick trip to Google with a search term of "how much uranium per petawatt" turns up a hit that does not answer the question directly but does tell you that it is a large mass.
Hey so I got another question. Is it correct for me to divide the energy in Joules equivalent to 2.49 PWh on 0.35 as it stands for 35% efficiency for the power plant?
 
  • #12
eminem4002 said:
Hey so I got another question. Is it correct for me to divide the energy in Joules equivalent to 2.49 PWh on 0.35 as it stands for 35% efficiency for the power plant?
If I was to multiply the value then it would just tell me what 35% of that energy is.
 
  • #13
eminem4002 said:
Hey so I got another question. Is it correct for me to divide the energy in Joules equivalent to 2.49 PWh on 0.35 as it stands for 35% efficiency for the power plant?
Yes. You start with the energy that was delivered by the plant and need to wind up with a larger number to reflect the energy in the uranium consumed by the plant. Multiplying by 0.35 would be wrong. Dividing was the correct thing to do.
 
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  • #14
eminem4002 said:
... However does it seem logical that it would take so many kilograms of uranium 235 to create 2.49 Peta Watthours? I know that peta is a huge number but would it really take so much mass to create this much electricity?
Are you sure the original number and units are correct?
There is no single power plant that could generate 2.49 petawatt-hour.

Copied from
https://www.iadb.org/en/news/latin-...r-electricity-needs-using-renewable-resources

"One petawatt-hour is equivalent to 1 trillion kilowatt-hour, roughly 3 times the amount of electricity Mexico consumes in one year.
At present, Latin America generates 1.3 petawatt-hour.
By 2050, demand is expected to grow to between 2.5 to 3.5 petawatt-hour."


Even one terawatt-hour, which equals one trillion (##10^{12}##) watts-hour, is used to express annual electricity generation for entire countries.
 
  • #15
Lnewqban said:
Are you sure the original number and units are correct?
There is no single power plant that could generate 2.49 petawatt-hour.

Copied from
https://www.iadb.org/en/news/latin-...r-electricity-needs-using-renewable-resources

"One petawatt-hour is equivalent to 1 trillion kilowatt-hour, roughly 3 times the amount of electricity Mexico consumes in one year.
At present, Latin America generates 1.3 petawatt-hour.
By 2050, demand is expected to grow to between 2.5 to 3.5 petawatt-hour."


Even one terawatt-hour, which equals one trillion (##10^{12}##) watts-hour, is used to express annual electricity generation for entire countries.
The task is completely theoretical, it said that the whole nuclear power plant industry had made 11,5% of the whole production of electricity which is equivalent to 2.49 PWh (in 2016).
 
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1. How does a nuclear power plant produce electricity?

A nuclear power plant produces electricity through a process called nuclear fission. This involves splitting atoms of uranium, which releases heat energy. The heat is then used to create steam, which turns a turbine and generates electricity.

2. Is nuclear power a safe way to produce electricity?

Nuclear power plants are designed with multiple safety features to prevent accidents and mitigate any potential risks. While there have been a few high-profile accidents in the past, overall, nuclear power is considered a safe and reliable source of electricity.

3. What happens to nuclear waste from power plants?

Nuclear waste from power plants is typically stored in special containers and then stored in a secure facility. Some countries also have plans to reprocess or recycle nuclear waste to reduce its volume and make it less hazardous.

4. How does nuclear power compare to other sources of electricity?

Nuclear power is a highly efficient and reliable source of electricity, producing large amounts of energy with minimal carbon emissions. However, it does have higher upfront costs and concerns about waste management and safety.

5. Are there any alternative uses for nuclear power plants besides electricity production?

Yes, nuclear power plants can also be used to produce heat for industrial processes, desalinate water, and even power space exploration. Some research is also being done on using nuclear power for propulsion in vehicles.

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