6 ms·

>shut off the pumps and radioactive decay alone can be enough to eventually cause a melt down in many modern designs The newest designs being built worldwide u

>shut off the pumps and radioactive decay alone can be enough to eventually cause a melt down in many modern designs The newest designs being built worldwide use natural circulation cooling and do not need cooling pumps in emergencies. Eventually the cooling pool needs to be refilled, but it's external to the containment pressure boundary, so you could refill it with a fire truck.

Retric 5y ago
Natural circulation gets heat from a reactor to X, but now your dependent on X. This often seems like a trivial detail, but Fukushima failed 3 days after the earthquake. The issue is you want several things from a passive system at the same time, don’t lose heat in normal operation, quickly lose multiple GW of heat in an emergency and as much as 200+MW of heat for days after a shutdown. The obvious solution is to have a tank of water that boils if the reactor temperature gets to high, but now you need to keep that tank full. Thus many designs result in a reactor that is passively safe for some number of hours and at risk after that. They describe this as a passively safe reactor even if it’s got external dependencies.
- i_am_proteus 5y ago
  Nope. Decay heat is below a half percent of operating after about a day. So 200 MW decay heat days later would mean a 40GW (thermal) reactor. That's about ten times larger than the largest reactors in existence today. Also, watts measure power, not heat.
  - Retric 5y ago
    Watts is joules per second be that electricity, horsepower, or heat. Passive systems can’t assume a successful shutdown.
    - mlyle 5y ago
      I can't figure out what you're saying here. You're assuming that the control rods aren't in, days later, for the purpose of calculating heat days after shutdown?
      - Retric 5y ago
        I am saying passive safety can’t assume anything else worked.
        mlyle 5y ago
        If you're going to apply "passive safety" globally, sure. But what we're talking about here is passive cooling system safety, not that the entire reactor is passively safe. The multiply-redundant shutdown systems suffice to end the chain reaction. If the chain reaction ends, you're pretty much immediately at 7% of decay heat -- so sure, a 1.5GW reactor will put out 100MW of decay heat, still. But this will rapidly fall off. After about an hour, it's more like 15MW; after a day, 6MW. Your statement of "200MW of decay heat" days later assumes either a ridiculous initial condition (an implausibly large reactor) or assumes you still have an operating reactor, which... isn't decay heat anymore.
        Retric 5y ago
        There have been multiple cases where reactors haven’t fully shut down safely. Assuming a scram will 100% work every time in an emergency simply isn’t appropriate or realistic. Also, Palo Verde Nuclear Generating Station has 3 different 4000MW thermal reactors, 7% of that is 280MW, though sure if everything shuts down properly it should hit ~15MW. Mitsubishi APWR is aiming for 4.5GW thermal in normal operation though some safety margin needs to be considered on top of that.
        mlyle 5y ago
        > There have been multiple cases where reactors haven’t fully shut down safely There's Chernobyl, and a few cases where a scram was delayed by 15 minutes or less. How can this produce hundreds of megawatts days later? > There have been multiple cases where reactors haven’t fully shut down safely. Assuming a scram will 100% work every time in an emergency simply isn’t appropriate or realistic. Assuming that every system fully fails is unrealistic, too. > 7% of that is 280MW, You said hundreds of megawatts days later. > Mitsubishi APWR is aiming for 4.5GW thermal in normal operation though some safety margin needs to be considered on top of that. What, they're going to run it over nameplate for days straight? A small excursion over 4.5GW won't appreciably change the amount of power output days later. Now you're just being silly.
    - i_am_proteus 5y ago
      The watt is a measure of power, heat (or energy) per unit time.
petre 5y ago
One could use pumps for increased efficiency during normal operation but the idea is that natural circulation should be able to remove all the heat if the reactor is SCRAMed. NuScale's design for instance only uses pumps for the steam generator, the rest is handled by natural circulation and the reactor sits in a water pool that needs to be replenished after two weeks in case of a major accident.
mlyle 5y ago
> so you could refill it with a fire truck. Note you might need a lot of fire trucks, though. If decay heat is 0.2% of nominal power after a week, we need .8mL per nameplate MW (thermal) per second... which doesn't sound bad, but a 2GW reactor needs 5000L firetrucks showing up every 50 minutes, 24/7, after a week. This could be challenging depending upon the underlying emergency. ((0.002 megawatt) / ((2260 + 250) * (kilojoules / kg))) / (997 (kg / (m^3))) = 0.00079921038 liters per second Assumption is that you're raising the temperature of the water 60C and then evaporating it, and that all the heat energy goes into the water. In the real world it can be expected to be slightly better than this, but not much...
- xyzzyz 5y ago
  You're off in your calculation by 3 orders of magnitude, you meant to write 0.002 gigawatt, not megawatt. Then yes, you get 0.7 liters/second, and 5000 L/hour for 2 GW plant. However, in actual reality, the firetruck would just park there, put one end of the hose into the plentiful source of water that surely must be available next to a power plant, and the other one into the entrance for coolant, and start its pump. Pumping 1 liter/second is quite in range of firetruck pumps abilities.
  - Retric 5y ago
    That’s a great assumption unless loss of water is the underlying emergency and to really mess things up you lost both water and the scram failed. That risk can be minimized by placing them near large natural bodies of water, but not all are ex: https://ejatlas.org/conflict/metsamor-nuclear-power-plant https://ejatlas.org/conflict/metsamor-nuclear-power-plant
  - mlyle 5y ago
    > You're off in your calculation by 3 orders of magnitude, you meant to write 0.002 gigawatt, not megawatt. No... decay heat is about .2% of nameplate power after a week. So for each megawatt of nameplate thermal power, you need to get rid of 2 kilowatts of heat-- or 0.8mL/second of water boiling off. "If decay heat is 0.2% of nominal power after a week, we need .8mL per nameplate MW (thermal) per second..." Was quite clearly said. > put one end of the hose into the plentiful source of water that surely must be available next to a power plant, and the other one into the entrance for coolant, and start its pump. Pumping 1 liter/second is quite in range of firetruck pumps abilities. This is pretty optimistic in many disaster scenarios and doesn't apply to all plants.

Retric 5y ago

Natural circulation gets heat from a reactor to X, but now your dependent on X. This often seems like a trivial detail, but Fukushima failed 3 days after the earthquake. The issue is you want several things from a passive system at the same time, don’t lose heat in normal operation, quickly lose multiple GW of heat in an emergency and as much as 200+MW of heat for days after a shutdown. The obvious solution is to have a tank of water that boils if the reactor temperature gets to high, but now you need to keep that tank full. Thus many designs result in a reactor that is passively safe for some number of hours and at risk after that. They describe this as a passively safe reactor even if it’s got external dependencies.

i_am_proteus 5y ago
Nope. Decay heat is below a half percent of operating after about a day. So 200 MW decay heat days later would mean a 40GW (thermal) reactor. That's about ten times larger than the largest reactors in existence today. Also, watts measure power, not heat.
- Retric 5y ago
  Watts is joules per second be that electricity, horsepower, or heat. Passive systems can’t assume a successful shutdown.
  - mlyle 5y ago
    I can't figure out what you're saying here. You're assuming that the control rods aren't in, days later, for the purpose of calculating heat days after shutdown?
    - Retric 5y ago
      I am saying passive safety can’t assume anything else worked.
      - mlyle 5y ago
        If you're going to apply "passive safety" globally, sure. But what we're talking about here is passive cooling system safety, not that the entire reactor is passively safe. The multiply-redundant shutdown systems suffice to end the chain reaction. If the chain reaction ends, you're pretty much immediately at 7% of decay heat -- so sure, a 1.5GW reactor will put out 100MW of decay heat, still. But this will rapidly fall off. After about an hour, it's more like 15MW; after a day, 6MW. Your statement of "200MW of decay heat" days later assumes either a ridiculous initial condition (an implausibly large reactor) or assumes you still have an operating reactor, which... isn't decay heat anymore.
        Retric 5y ago
        There have been multiple cases where reactors haven’t fully shut down safely. Assuming a scram will 100% work every time in an emergency simply isn’t appropriate or realistic. Also, Palo Verde Nuclear Generating Station has 3 different 4000MW thermal reactors, 7% of that is 280MW, though sure if everything shuts down properly it should hit ~15MW. Mitsubishi APWR is aiming for 4.5GW thermal in normal operation though some safety margin needs to be considered on top of that.
        mlyle 5y ago
        > There have been multiple cases where reactors haven’t fully shut down safely There's Chernobyl, and a few cases where a scram was delayed by 15 minutes or less. How can this produce hundreds of megawatts days later? > There have been multiple cases where reactors haven’t fully shut down safely. Assuming a scram will 100% work every time in an emergency simply isn’t appropriate or realistic. Assuming that every system fully fails is unrealistic, too. > 7% of that is 280MW, You said hundreds of megawatts days later. > Mitsubishi APWR is aiming for 4.5GW thermal in normal operation though some safety margin needs to be considered on top of that. What, they're going to run it over nameplate for days straight? A small excursion over 4.5GW won't appreciably change the amount of power output days later. Now you're just being silly.
  - i_am_proteus 5y ago
    The watt is a measure of power, heat (or energy) per unit time.

petre 5y ago

One could use pumps for increased efficiency during normal operation but the idea is that natural circulation should be able to remove all the heat if the reactor is SCRAMed. NuScale's design for instance only uses pumps for the steam generator, the rest is handled by natural circulation and the reactor sits in a water pool that needs to be replenished after two weeks in case of a major accident.

mlyle 5y ago

> so you could refill it with a fire truck. Note you might need a lot of fire trucks, though. If decay heat is 0.2% of nominal power after a week, we need .8mL per nameplate MW (thermal) per second... which doesn't sound bad, but a 2GW reactor needs 5000L firetrucks showing up every 50 minutes, 24/7, after a week. This could be challenging depending upon the underlying emergency. ((0.002 megawatt) / ((2260 + 250) * (kilojoules / kg))) / (997 (kg / (m^3))) = 0.00079921038 liters per second Assumption is that you're raising the temperature of the water 60C and then evaporating it, and that all the heat energy goes into the water. In the real world it can be expected to be slightly better than this, but not much...

xyzzyz 5y ago
You're off in your calculation by 3 orders of magnitude, you meant to write 0.002 gigawatt, not megawatt. Then yes, you get 0.7 liters/second, and 5000 L/hour for 2 GW plant. However, in actual reality, the firetruck would just park there, put one end of the hose into the plentiful source of water that surely must be available next to a power plant, and the other one into the entrance for coolant, and start its pump. Pumping 1 liter/second is quite in range of firetruck pumps abilities.
- Retric 5y ago
  That’s a great assumption unless loss of water is the underlying emergency and to really mess things up you lost both water and the scram failed. That risk can be minimized by placing them near large natural bodies of water, but not all are ex: https://ejatlas.org/conflict/metsamor-nuclear-power-plant https://ejatlas.org/conflict/metsamor-nuclear-power-plant
- mlyle 5y ago
  > You're off in your calculation by 3 orders of magnitude, you meant to write 0.002 gigawatt, not megawatt. No... decay heat is about .2% of nameplate power after a week. So for each megawatt of nameplate thermal power, you need to get rid of 2 kilowatts of heat-- or 0.8mL/second of water boiling off. "If decay heat is 0.2% of nominal power after a week, we need .8mL per nameplate MW (thermal) per second..." Was quite clearly said. > put one end of the hose into the plentiful source of water that surely must be available next to a power plant, and the other one into the entrance for coolant, and start its pump. Pumping 1 liter/second is quite in range of firetruck pumps abilities. This is pretty optimistic in many disaster scenarios and doesn't apply to all plants.