# From the Fukushima Daiichi forum - heating and cooling with water

1. Mar 5, 2012

### duccio

Hi folks, this topic originated in the Fukushima Daiichi forum here on PhysicsForum, but since it involves basic physics math I was doing almost 15 years ago in college, I figured out here would be a better place to work out a solution

1. The problem statement, all variables and given/known data

The melted core of reactor II produces 0.9 megawatts of decay heat (about 10 kW/ton after one year, fission heat is now negligible). Tepco cools it with 8.9 m3/hour of water. Inflow water temperature should be around 10 degree (from the weather forecast in Japan). Tepco states that everything inside the reactor is <100 degree Celsius, more precisely at around 43 degree, at atmospheric pressure.

So we want to do a little bit of investigation, and answer the followings:

1) How many joule/hour are 900KW?
2) Let's make it simple: supposing all and only the water injected in one hour goes on top of the fuel, how many joules per cubic centimeter of water of heat is transferred to the water?
3) What would be the temperature of the water at the end of the hour? (the deltaT)

The end point is to determine whether Tepco statements are plausible: Is 8.9t/hour of water sufficient for cooling the core of the reactor at 45 degree?

2. Relevant equations

W = J/s , or simplier, 1KWh = 3.6MJ
1 Calorie (small) = around 4.2 J

3. The attempt at a solution

1) 900KWh * 3.6MJ = 3.24GJ
2) 1 gram of water is approximately 1 cubic centimeter. 8.9 m3 water are 8.9M cc. 3.24GJ divided by 8.9M cc = 364 Joule/gram of water
3) 364Joule/gram is around 87 calories, or +87 degree. With 10 degree from the source, the final temperature of the water at the end would be 97 degree.

If the math is right, the inside of the reactor shouldn't be closer to 100 degree than to 45?

Duccio
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution