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NHK World English reports this story: Cooling system operating at No. 2 reactor

The operator of the troubled Fukushima Daiichi nuclear power plant has started operating a system to effectively cool water in a spent fuel pool in the plant's No.2 reactor building.

The Tokyo Electric Power Company on Tuesday set up at the building the first circulatory cooling system to be installed at the plant since the accident in March.

The utility has been pumping about 50 tons of water into the pool every few days.

The pool's temperature is around 70 degrees Celsius, apparently producing steam that has filled the building and resulted in a humidity level of 99.9 percent.

The humidity and high radiation levels have been hampering repair work at the site.

The new system is to pump water out of the pool to a heat exchanger and return the water to the pool as coolant.

The firm says it plans to lower the pool's temperature to around 40 degrees Celsius in a month and hopes to reduce the humidity level before installing equipment to remove radioactive substances in the building.

The firm says it will start operating similar systems at the plant's No.1 and 3 reactors in June, and at the No.4 reactor in July.

Tuesday, May 31, 2011 20:57 +0900 (JST)

Can anybody see what the problem with this report is? Plain physics is the problem…

1. If “The utility has been pumping about 50 tons of water into the pool every few days.” where is the water going? Fifty tons of water, assuming metric tons because we are in Japan, equates to 100,000 kg of water, which equates to ~26,000 gallons, or ~100,000 liters. That's a lot of water, EVERY FEW DAYS. Where is it going? If they are talking English units, 50 tons of water is 100,000 lb of H₂O, which equates to ~12,000 gallons of water or ~45,000 liters. Still a lot of water to be feeding into the cooling pool at Daiichi II.
    
2. Water boils at 100°C. Water at 70°C will create water vapor, but not steam. True, the building WILL BE a sauna, but the problem here is that water at 70°C will not evaporate away 26,000 gallons of water in 24 hours. So we are back to the issue of where is all of that water going that is being pumped in?

   If you set a gallon of water on the stove, and set the burner so that the temperature of the water is 70°C (160°F), how long will it take that gallon to evaporate? See the surface evaporation calculator at EngineeringToolbox to try some of the values that are possible. They values don't work out... Different parameter values give between 0.013 - 0.017 kg/s evaporation rate, which times 24 hours gives 1100 kg - 1500 kg of water evaporating. Maybe the parameters used in the equations are wrong? The above values were for a 100 m² surface area at 70°C with 100% humidity, and 0 m/s airflow. Using 282 gm/kg of water in the air at saturation and 3 to 10 gm/kg of water at 0% humidity. Smaller surface areas will give smaller evaporation rates.

Something doesn't jive, and it has lead to my swearing off of sushi for probably the rest of my life…