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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
U,
contribution from
although comparable to that from Pu, would still represent an insignificant source of
56

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