3 ms·
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 f
by 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 agoThat’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.