Question 3

The radioisotope battery is fuelled with 500g of Pu²³⁸C (Plutonium-238 carbide), which has a density of 12.5 g/cm³. The Pu²³⁸ has a half life of 89 years, and emits 5.6 MeV per disintegration, all of which may be assumed to be absorbed in the generator. The thermal to electrical efficiency of the system is 6 percent. Calculate (a) the specific power in watts (thermal) per gram of fuel; (b) the power density in watts (thermal) per cm³; (c) the fuel efficiency in curies per watt (thermal); (d) the total electrical power of the generator.

Solution 3

λ =  0.693/T_1/2

    = 0.693/(89 x 365.25 x 24 x 3600)

    = 2.467 x 10^-10 s^-1

Molecules/unit mass = N = 6.023 x 10²³/250

                                        = 2.409 x 10²¹ g^-1

Specific activity = λN = 2.467 x 10^-10 x 2.409 x 10²¹

                                    = 5.944 x 10¹¹ s^-1/g

                                    = 5.944 x 10¹¹ /3.7 x 10¹⁰

                                     = 16.07 Ci/g

Specific Power= λNE = 5.944 x 10¹¹ x 5.6 x 1.603 x 10^-13

                                   = 0.534 W/g

Power Density = λNEρ = 0.534 W/g  x  12.5 g/cm³

                                      =  6.67 W/cm³ 

Fuel Efficiency = λN/ λNE = 16.07 Ci/g  / 0.534 W/g

                                           =  30.11  Ci/W 

Total Power = Specific power  x Mass x Conv. Eff.

                     =λnEmη = 0.534 W/g x 500 g x 0.06

                     =  16.01  W(electric)