3 ms·

About the water vaporizing: I'm guessing the pool is so "huge" that the sheer amount of energy required to vaporize the water can't be found in the fissile fuel

by vages 7y ago

About the water vaporizing: I'm guessing the pool is so "huge" that the sheer amount of energy required to vaporize the water can't be found in the fissile fuel alone. That doesn't account for the possibility of a leak, though.

rwcarlsen 7y ago
In an accident scenario, the heat that is problematic is not from the fissile material (i.e. U235, Pu239) but from the fission products' decay heat - the isotope soup remaining after the fissile atoms split. It is much less intense than the heat/energy produced during normal operation and drops off significantly after a couple days. So the pool's purpose in terms of accident mitigation is to keep things cool enough during the period of time after reactor shutdown and before radioactive isotopes left over from split atoms decay/cool-down enough.
- petschge 7y ago
  While you are correct that it is less than the nominal heat output, we are still talking about 50MW a minute after shut down and 5MW one day after the shutdown of a 700MW reactor. So even a day after the shutdown you are able to boil of 2kg of water every second.
  - rwcarlsen 7y ago
    Yes, it is still a lot of heat. But to a lay reader, it might be easy to believe that the pool needing to be able to cool the reactor means that the pool has to cool an operating plant during an accident which is incorrect. I was just trying to make the distinction here. Also the drop from (your very approx numbers) 50 MW to 5 MW is mostly exponential - and so very quickly drops away from 50 MW. After that, any remaining heat is diffuse enough to not actually boil away water from the pool. Fuel-melting risk (which is not the chain reaction restarting for any who may be limited in fission plant understanding) is generally only a concern for the first couple days after reactor shutdown.
    - petschge 7y ago
      Funny little detail that might interest you: Due to the presence of a large number of different fission products (with different half lifes), the drop does not really follow an exponential but is better described by P(t) = P_0 * a * (t^-0.2), i.e. a power law, where P_0 is the power before the shutdown, a is a couple of percent and t is the time since the shutdown in seconds.

by vages 7y ago

rwcarlsen 7y ago

In an accident scenario, the heat that is problematic is not from the fissile material (i.e. U235, Pu239) but from the fission products' decay heat - the isotope soup remaining after the fissile atoms split. It is much less intense than the heat/energy produced during normal operation and drops off significantly after a couple days. So the pool's purpose in terms of accident mitigation is to keep things cool enough during the period of time after reactor shutdown and before radioactive isotopes left over from split atoms decay/cool-down enough.

petschge 7y ago
While you are correct that it is less than the nominal heat output, we are still talking about 50MW a minute after shut down and 5MW one day after the shutdown of a 700MW reactor. So even a day after the shutdown you are able to boil of 2kg of water every second.
- rwcarlsen 7y ago
  Yes, it is still a lot of heat. But to a lay reader, it might be easy to believe that the pool needing to be able to cool the reactor means that the pool has to cool an operating plant during an accident which is incorrect. I was just trying to make the distinction here. Also the drop from (your very approx numbers) 50 MW to 5 MW is mostly exponential - and so very quickly drops away from 50 MW. After that, any remaining heat is diffuse enough to not actually boil away water from the pool. Fuel-melting risk (which is not the chain reaction restarting for any who may be limited in fission plant understanding) is generally only a concern for the first couple days after reactor shutdown.
  - petschge 7y ago
    Funny little detail that might interest you: Due to the presence of a large number of different fission products (with different half lifes), the drop does not really follow an exponential but is better described by P(t) = P_0 * a * (t^-0.2), i.e. a power law, where P_0 is the power before the shutdown, a is a couple of percent and t is the time since the shutdown in seconds.