Decay Helium - Continued - std301320040064

DOE-STD-3013-2004

where E is the energy emitted/generated during radioactive decay. For simplicity, consider only

the initial heat generation. Then solve for N0, substitute the result into the equation for helium

produced, and use the linear approximation for the exponential, yielding

H = Q0τ/E.

[21]

Using this relationship, the pressure equation becomes

PHe = 14.7x22.4(Q0τ/E)(T1/273)/V1

PHe = 1.2061(Q0τ/E)T1/V1

[22]

Note that for Q0 in watts and τ in years, the quantity E must be in watt-years/mole. Table B-5

(found in Section B.4.1) provides decay energies for radionuclides of interest in both Mev and

watt-yr/mole. As can be seen from the values in that table, Equation [22] produces pressures

that are relatively insensitive to the radioactive species chosen because the value of E is

relatively constant over the species considered. The conservative evaluation is achieved by

239

Pu. When that is

using a relatively low value for E. A reasonable selection is the value for

used, Equation [22] becomes

PHe = 7.517x10-5 Q0τT1/V1

[23]

Note that Q0 is the Specific Heat Generation Rate (SHGR) (from the last column of Table B-5)

239

Pu (1.93 w/kg) is

times the mass of plutonium in the container. If the value in Table B-5 for

used, and a conversion to use kg of oxide (instead of kg of plutonium) is also made to give the

relationship Q0 = 1.7 m, the resulting equation is the same as the one above based on mass

(Equation [19]):

PHe = 7.517x10-5 (1.7 m)τT1/V1

= 1.28x10-4 mτT1/V1

Note also that the contribution from decay of uranium isotopes is negligible, with the possible

233

U. As an extreme case, consider an oxide material with a composition of

exception of

235

239

233

U, 0.1 wt%

Pu, and 0.5 wt%

U. In such a material, the

approximately 88 wt%

235

233

U would be only about 3 % of the Pu contribution, and that from

contribution from

although comparable to that from Pu, would still represent an insignificant source of