TO:

FROM:

SUBJECT: Heat from Fission Product Radioactivity in Reactor

CC: Reactor Committee

answer the question very nicely.

The 10 Kw water boiler operates continuously up to a shutdown. It is

sary

Reference: | 1. Goodman, Vol. I, p.243 |

2. K. Way, Phys. Rev. 70, 115 (1946) |

K. Way gives the following expression for energy dissipation from fission products

by [beta] and [gamma] radiation.

E[beta] and [gamma] = 2.66 t-1.2 Mev/sec-fission | 10<t<107 seconds | |

(goodman) | (within factor of 2) | |

Solution: |

The contribution to power at time t by products produced in the interval dT is:

The total is the integral:

In terms of watts:

If the reactor has been operating for say 6 months, then To in seconds is 6 x 30 x

86,400 = 15,550,000 seconds.

Evaluating E at t = 10 seconds:

at t=100 seconds: |

Conclusions:

At 100 seconds after shutdown, the maximum [beta] and [gamma] power from decay of fission

products is 500 watts assuming:

- 1. K Way's formula is incorrect adversely by a factor of 2.
- 2. All [beta] and [gamma]'s are absorbed in reactor.

Since 1/2 Kw is removed from reactor by the surroundings, the solution does not

boil.

That this estimate is reasonable may be seen by the following considerations.

(Goodman,I, p. 240) Of the 200 Mev produced by a fission, ~ 22 Mev is due to

fission product decay. "About half of this last figure is emitted as neutrino

energy." Therefore, only ~11 Mev is available as [gamma] and [beta] energy from fission

products. That is, if equilibrium is nearly established between the fission

products and a 10 Kw power level, then the steady state decay contribution will be

approximately 11/200 x 10 Kw = 550 watts. This decays rapidly upon reactor shutdown

and represents a maximum "decay power" in a reactor not exceeding 10 Kw.

Also, the 1/2 Kw conduction by surrounds is conservative in that an increase

in fuel temperature above normal operation increases the heat removed by

ing

is, if normally the conditions are 85°-25°C removing .5 Kw of heat, then at

100°-25°C the heat removal will be ~ 75/60 x .5 = .625 Kw.

This is the maximum heat removal by surroundings if the boiling point is

100°C. However, the boiling point will be higher due to the involatile solute and

may be computed by an elementary method given in "Outlines of Physical Chemistry"

by

If there is some question about the closeness of the power production to

heat conduction shortly after shutdown it may be shown that if the contents of the

water boiler is assumed to be 1 ft³ of water, then a net gain of 500 watts for

a period of almost one hour is required to raise the temperature to boiling. At

one hour