Formula of decay heat for a fraction of radioactive materials

Question

What is the formula of decay heat for a fraction of spent fuel? (say 50% spent fuel)

The Wigner-Way formula has not any parameter relating to the amount of fuel. Is there any other formula?

Answer 1

The page you link says

It is also possible to make a rough approximation by using a single half-life that represents the overall decay of the core over a certain period of time. An equation that uses this approximation is the Wigner-Way formula: $$ P_d(t) = 0.0622 P_0 \left( t^{-0.2} - (t_0 + t)^{-0.2} \right) $$

where

• $P_d(t)$ is thermal power generation due to beta and gamma rays,
• $P_0$ is thermal power before shutdown,
• $t_0$ is time, in seconds, of thermal power level before shutdown,
• $t$ is time, in seconds, elapsed since shutdown.

The emphasis on “rough approximation” is in the original. What happens is that, during fission, short-lived fission products with lots of different lifetimes accumulate towards a secular equilibrium. When power generation stops, the short-lived stuff dies away quickly and the long-lived stuff stays on. If you have unspent fuel ($t_0 = 0$), this approximation says that the decay heat is always zero.

It’s not really correct to refer to this as “a single half-life” as your link does. A single half-life would go like $2^{-t/\tau_\text{single}}$. This relationship says that, when you add up all of the half-lives of all the messy garbage that comes out of uranium fission, a time dependence of $t^{-1/5}$ captures the way that the decay radiation dies off more rapidly at the beginning than later on. It’s an empirical relationship.

You write,

The Wigner-Way formula has not any parameter relating to the amount of fuel.

But it does: the amount of spent fuel is related to the thermal power $P_0$ while the reactor was operating. The decay heat depends on how much fission product you have, not how much fuel you started with. Possibly related.

In a comment you ask

Can you please tell me what is the 0.006 in the Wigner-Way formula?

There isn’t one. There is a numerical constant that’s ten times bigger than that, however.

A physicist who’s accustomed to doing dimensional analysis would be tempted to look at that factor and conclude that the decay power starts off as 6% of the thermal power. But that’s bogus, because the empirical Wigner-Way formula has screwy units. That factor $0.0622$ has dimension $\text{(seconds)}^{+0.2}$, and the exponent $+0.2$ has more to do with fitting a bunch of exponentials together that with any nice interpretable algebra. If you were to do the same kind of approximation for plutonium fuel, which has a different spectrum of short-lived decay products, you might expect a different exponent.

Answer 2

There can be no general formula. The energy released depends on the radioactive decay of each radionuclide taken one by one. See what are the " decay schemes " for each radionuclide, for example for 60Co. It is different for 137Cs . Now , if you think " nuclear fuel" like 235U , one atom gives one fission which products approx 200 MeV . All radioactive materials are not " nuclear fuel " .